I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

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    $\begingroup$ Do they know about logarithms? $\endgroup$
    – donaastor
    Feb 27, 2021 at 19:35
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    $\begingroup$ Maybe try this. Firstly, restrict $x$ to natural numbers. Now take the ratio of two consecutive elements in each of your two arrays. Exponential function will have constant ratio $1.0001>1$, while the ratio of quadratic expression will converge to $1$. Say that it happens at $x=n$ that the ratio of $1000x^2$ drops below $1.0001$. After that happens, you just have to explain why an exponential function (this time some other than $1.0001^x$) will eventually be greater than arbitrary constant. That constant will be $\frac{1000n^2}{1.0001^n}$. $\endgroup$
    – donaastor
    Feb 27, 2021 at 19:41
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    $\begingroup$ If you observe the "difference pyramid", the quadratic (e.g. $x^2$) has two more layers $$ [0, 1, 4, 9, 16, 25, 36, 49, 64, 81, \dots]\\ [1, 3, 5, 7, 9, 11, 13, 15, 17, \dots]\\ [2, 2, 2, 2, 2, 2, 2, 2, \dots]\\ $$ but the exponential (e.g. $3^x$) doesn't have a bottom. $$ [1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, \dots]\\ [2, 6, 18, 54, 162, 486, 1458, 4374, 13122, \dots]\\ [4, 12, 36, 108, 324, 972, 2916, 8748, \dots]\\ [8, 24, 72, 216, 648, 1944, 5832, \dots]\\ [16, 48, 144, 432, 1296, 3888, \dots]\\ \dots $$ $\endgroup$
    – Vepir
    Feb 27, 2021 at 19:49
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    $\begingroup$ Well, $$e^x=(e^{x/2})^2=(e^{x/3})^3=(e^{x/4})^4=(e^{x/5})^5=\ldots,$$ so if it's bigger than a polynomial of degree $d$, it's eventually bigger than a polynomial of degree $2\,d, 3\,d, 4\,d, 5\,d,\ldots$. That's why we call it "growing exponentially". $\endgroup$
    – NoNames
    Feb 27, 2021 at 19:56
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    $\begingroup$ @peek-a-boo: To someone in 7th grade, what might be considered a hand-wavey explanation to a mathematician would seem like a perfectly valid justification. $\endgroup$
    – Joe
    Feb 27, 2021 at 22:40

15 Answers 15


If you have a quadratic polynomial $f(x)$ and an exponential function $b(x) = b^x$ where $b>1$, you can show that $b(x)$ surpasses the polynomial by showing that eventually the growth rate of $b(x)$ exceeds the polynomials growth rate.

Since the polynomial's leading term has the biggest effect when $x$ grows very big, that means that the other terms don't matter, so let $f(x) = ax^2$ for an $a>0$. Now, compute the ratio between $f(x+1)$ and $f(x)$: $$ \frac{f(x+1)}{f(x)} = \frac{a(x+1)^2}{ax^2} = \left(\frac{x+1}{x}\right)^2 \, . $$ As you can see, as $x$ grows very big, that ratio between $x+1$ and $x$ grows close to $1$, thus $f(x+1)$ barely increases from $f(x)$ because it is being multiplied by a number close to $1$. Now analyze the ratio between $b(x+1)$ and $b(x)$. By definition, the exponential function multiplies by its base $b$ evey time you increase by $1$, so the ratio between $b(x+1)$ and $b(x)$ is always $b$. However, we stated already that $b>1$. We also found out that $f(x)$ ratio approaches $1$ as $x$ gets really big. Thus, there is a point when $b(x)$ ratio exceeds $f(x)$ ratio, which means that $b(x)$ will start growing faster than $f(x)$ and will eventually outgrow $f(x)$.

Note that I just used terminology like approach and really big because a $7$th grader would not know of limits, so don't nitpick that.

Secondary note: I said $a>0$ and $b>1$ because I assumed that both of the functions would be traveling upwards as you moved right along the $x$-axis.

If you want a downward-facing parabola with $a<0$ and a downward-facing exponential with $0<b<1$, then you can just note that the exponential will just tend towards $0$ when $x$ gets very big, but the quadratic will eventually go below zero if it is facing downwards, thus showing that the exponential will eventually become greater then the quadratic.

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    $\begingroup$ That's an interesting approach to bringing differential calculus to the problem in a way that a 7th grader can probably understand. $\endgroup$ Feb 28, 2021 at 14:08
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    $\begingroup$ Incidentally, it's a very short path from what you are describing to big O notation. Big O is fantastically useful in computer science, but it can lead to bad habits in high school/undergraduate calculus (e.g. when asked to simplify $\lim_{x\to\infty} \frac{5x}{x+3}$, you "just ignore the 3" and write 5), so it's best consumed in moderation at that level. $\endgroup$
    – Kevin
    Mar 2, 2021 at 1:39
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    $\begingroup$ @Kevin What is wrong with that approach, it has been very helpful for me $\endgroup$
    – Some Guy
    Mar 2, 2021 at 1:43
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    $\begingroup$ @Kevin The reason you "just ignore the 3" is actually straightforward. $lim_{x\to\infty}\frac{5x}{x+3}$ = $5 \lim_{x\to\infty}\frac{x}{x+3}$ = $5 \lim_{x\to\infty}\frac{\frac{x}{x}}{\frac{x}{x}+\frac{3}{x}}$ = $5 \lim_{x\to\infty}\frac{1}{1+\frac{3}{x}}$ = $5 \lim_{x\to\infty}\frac{1}{1+\infty}$ = 5 $\endgroup$
    – mazunki
    Mar 2, 2021 at 20:14
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    $\begingroup$ I realize I wrote 1 over $1+\infty$ in the last step, but that's 1+0, leaving the limit being 1. Can't edit it anymore. $\endgroup$
    – mazunki
    Mar 2, 2021 at 20:20

For one example, look at how $2^x$ grows in comparison with $x^2$. If $f(x)=2^x$, then $f(2x)=2^{2x}=(2^x)^2$. So doubling the input means that the output is squared (!). By contrast, if $g(x)=x^2$, then $g(2x)=(2x)^2=4x^2$, and so the output is quadrupled. Ultimately, squaring is much more powerful than quadrupling, and so the exponential function grows faster.

  • 15
    $\begingroup$ This is a neat approach, since it reduces "exponentials dominate polynomials" to "quadratic functions dominate linear functions". $\endgroup$
    – Ian
    Feb 27, 2021 at 23:35
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    $\begingroup$ Thank you @Ian. It’s still far off a rigorous proof, but I hope that this explanation would be suitable for a 7th grader. $\endgroup$
    – Joe
    Feb 28, 2021 at 0:19

Assuming the exponential is $c^x$:

Then: $c^x = ((\sqrt{c})^2)^x = (\sqrt{c})^{2x} = ((\sqrt{c})^x)^2$

Our original question is now changed into: let's compare $x^2$ with $((\sqrt{c})^x)^2$. Which has the same result as comparing $x$ directly with $(\sqrt{c})^x$.

As a result, we have reduced our argument to: why does an exponential function eventually get bigger than a linear one? Which likely has been answered already by the teacher.

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    $\begingroup$ And the last question, since exponentiation is repeated multiplication and linear functions are repeated addition, boils down to "why is multiplication more than addition?", which anyone can understand. $\endgroup$ Mar 2, 2021 at 5:31

Personally I'd be hesitant to pose such a question to a $7$-th grader. For some this is a big but instructive stretch of the mind, for others the stretch is too big. So my first advice would be

Don't try to force your son to 'get' this.

Reservations aside, here's how I'd approach it: Together you can explore the question

How many steps does it take for such a function to double?

An exponential function grows by $c\%$ at every step. So after some fixed number of steps the output doubles; it isn't difficult to find that it has certainly doubled after $\tfrac{100}{c}$ steps.

This is of course clearer with an example; the output of the function $f(x)=1.02^x$ grows by $2\%$ every time $x$ increases by $1$. So if $x$ increases by $50$ then $f$ increases by at least $100\%$. That is to say $f$ more than doubles every $50$ steps. This is true regardless of the value of $x$! In the same way, every exponential function has some fixed number of steps required to double.

Conclusion 1: An exponential function has a 'doubling number'; every time you increase $x$ by this number, the function doubles.

How many steps do you need for the function $f(x)=Cx^2$ to double, for some constant $C$? This boils down to finding $y$ such that $f(y)=2f(x)$, i.e. $$Cy^2=2Cx^2,$$ and a bit of algebra then shows that $y=\sqrt{2}x$. So for small $x$ the function doubles quickly, but it takes longer and longer to double. For example, for $x=5$ you find $y\approx7$, so it takes only about $2$ steps for $f$ to double. But from $x=100$ it already takes roughly $41$ steps to double, and from $x=500$ it takes well over $200$ steps!

Conclusion 2: A quadratic function takes longer and longer to double as $x$ increases. The number of steps required to double increases without bound.

So eventually a given quadratic function will take longer to double than a given exponential function. From this point on the exponential function grows faster than the quadratic function, as it doubles faster. And so eventually the exponential function will overtake the quadratic function.

  • 2
    $\begingroup$ +1 Don't try to force your son to 'get' this. $\endgroup$
    – GEdgar
    Mar 2, 2021 at 18:37
  • $\begingroup$ Props for not forcing kids to learn stuff that ultimately isn't 'really needed for life'. Regardless, I especially enjoy the way this method first proves "for any combinatioon of functions, there is a number $x_0$, unknown, for which any bigger number than $x_0$ leaves $f(x_0)$ bigger than $g(x_0)$", and then just extrapolate that we don't really need to figure out what this number is for the statement to be true. $\endgroup$
    – mazunki
    Mar 2, 2021 at 20:09

Every time $x$ is incremented by $1,$ $3^x$ gets $3$ times as big.

When $x$ is incremented from $1\,000\,000$ to $1\,000\,001,$ $x^2$ increases by $2\,000\,001$ which may seem big, but is tiny by comparison to $1\,000\,000^2 = 1\,000\,000\,000\,000.$


Not sure they saw $\log$ yet, but to "visualize" and get the numbers out of the exponentiation, reduce the scale using the $\log$

$$\log(2^x) = x \cdot \log(2)$$ $$\log(x^2) = 2 \cdot \log(x)$$

Removing the constants, we compare now $x$ and $\log(x)$, and $$\dfrac{x}{\log(x)}$$grows quickly.


This question motivates thinking about exponents and hence logarithms. But even without going into the definition of logarithms, we may consider relevant expressions of $x$-values for which the difference in rate of growth may become apparent. Consider: $$ f(x)=a^x\\ g(x)=x^2 $$ and derive that $$ f(2x)=a^{2x}=\left(a^x\right)^2=g(a^x) $$ So now we can instead ask:

Does $f(2x)=g(a^x)$ eventually become larger than $g(2x)$?

So the problem now is reduced comparing the two inputs $a^x$ and $2x$ to the same function $g$ and showing that $a^x$ eventually becomes larger than $2x$ by any given factor one may wish to consider, however large.

As a corollary, this approach works for any degree polynomial, namely take $g(x)=x^n$ and consider $f(nx)=g(a^x)$ compared to $g(nx)$. Thus one compares an exponential input $a^x$ to a linear input $nx$.

Finally, in order to give the complete picture, why does $a^x$ exceed any linear term $kx$ eventually (given $a>1$)? Well, just consider: $$ f(x+1)-f(x)=a^{x+1}-a^x=a^x(a-1)\\ g(x+1)-g(x)=k $$ So now we only need to show that $a^x(a-1)$ eventually becomes larger than $k$ in order to see that an increase by $1$ in $x$ eventually will make $f$ grow at a faster rate than $g$. Clearly this expression $a^x(a-1)$ becomes larger an larger, and hence $f$ grows faster than a linear function with a greater slope than $g$ and will eventually catch up.


..it might be just me, but I think we all are reading this question wrong. This is 7th grade, so I don't think the teacher is asking for a specific proof of that assertion. The answer might be way simpler.

Notice that the question asks how you know, not why this happens.

"How do you know an exponential expression will be larger than any quadratic expression?"

The answer for that would be:

"An exponential expression will be larger than any quadratic expression when the base is larger than 1".

My rationale is that this "feels" a lot like "how do you know a number is divisible by 3?", to which the answer would be "when the sum of its digits is also divisible by 3".


Perhaps still the best to work on a concrete example: $1.0001^x$ vs. $1000x^2$ is a good one.

Can you prove that you can find big enough $x_0$ so that the ratio of $\frac{1000(x+1)^2}{1000x^2}$ becomes smaller than, say, $1.00005$ for all $x\ge x_0$? This ratio is $(1+\frac{1}{x})^2$ and one could perhaps try to solve the inequality $1+\frac{1}{x}\le\sqrt{1.00005}$. (Use the calculator to calculate the latter - you get that $x\ge x_0\approx 40,000.5$ would work.)

Now, from that point on, $1.0001^x$ may still be tiny compared to $1000x^2$, but whenever $x$ grows by $1$, $1.0001^x$ grows by a factor $1.0001$, while $1000x^2$ grows by a factor at most $1.00005$. Thus their quotient grows by a factor at least $\frac{1.0001}{1.00005}\approx 1.00005$. Even if it is very small to start with (near $x=x_0$), that quotient grows exponentially and will soon become bigger than $1$.

(Left as exercise: why does $1.00005^x$ eventually becomes bigger than any constant? Could use $1.00005^x=(1+0.00005)^x\ge 1+0.00005x$, the latter easily proven at least for natural $x$ by reasoning in an inductive way, even if mathematical induction has not been formally introduced.)

I don't know of a simpler proof, and my honest opinion is that this is probably not the topic suitable for introduction at that level. Maybe they have a very ambitious teacher who sadly cannot identify with the ways of thinking of the students at that level, and believes that, however amazing the mere fact is (that the exponential always grows faster than a polynomial), they can provoke more amazement by introducing the proof of that fact. I expect the effect to be quite the opposite - the kids may be put off maths for life. (Majestic however it is, the middle of Pacific ocean is not the place to drop a beginner swimmer to enjoy the beauty of the ocean, so to speak.)

So your best strategy may be to not do the assignment, and wait to see what the teacher would say. At least you spare your kid from the hours of frustration of them working with you, and the teacher may end up producing some handwavy argument in the end, which we will all accept as "good enough at this stage" - and move on.


I'd use a geometric argument combined with the principle of induction to convince your son. A simple, but fairly rigorous proof that is probably understandable to a middle schooler taking algebra.

Have your son imagine a square sheet of paper with side length $b$, where $b$ is very large, then by increasing its side length by $1$ we get a strip of paper added to the top and side of square, but increasing its dimension by $1$ will add far more material, since the strip of paper can clearly be cut and fit completely inside the cube with plenty of room to spare.

Therefore, $(b + 1)^2 < b^{2 + 1}$ (again, reminding your son that $b$ is very large).

In fact, if you like, you can just exhibit such a $b$ without giving the argument above, but I think it may help it sink in that increasing the variable in the exponential increases the volume of the square by an entirely new dimension while increasing the variable in a square just adds a tiny 2d slit of paper in comparison.

Now we just "repeat" the argument (you can use the word 'induction' if you like, but I think 'repeat' gets the idea across). Suppose that we know

$$(b + n)^2 < b^{2 + n}$$

for some $n$. We want to show that

$$(b + n + 1)^2 < b^{2 + n + 1}$$

we can rewrite this as

$$(b + n)^2 + 2(b + n) + 1 < b^{2 + n} b$$ The LHS rewrite can be seen geometrically. The side and top strips have length $(b + n)$ and height 1, and the corner has area 1, a diagram can be drawn to see this.

(Alternatively, not sure if they use this term in school anymore, but this way of rewriting the LHS was called "FOILing" when I was in middle school)

This inequality is definitely true, since

$$b^{2 + n} b = b^{2 + n} + \dots + b^{2 + n} \text{ ($b$ times)} $$,

This can also be argued geometrically, but it's trickier to visualize in dimensions higher than 3.

By our assumption, $$(b + n)^2 < b^{2 + n}$$

Therefore, $$(b + x)^2 < b^{2 + x}$$ for all natural numbers $x$ greater than or equal to 1.

If your son is interested in these types of things, then thinking about 'higher' dimensions will probably add an air of intrigue to the whole thing. The above induction step can be explained geometrically, and even still remain rigorous if a bit more work is put into it.

The idea is that, when you add 1 to the length of a square, you are adding only a 2 dimensional amount of stuff, but when you add 1 to the dimension of a square, you are adding a 3 dimensional amount of stuff.

When you add two to the length of the square, you are still adding only a 2 dimensional amount of stuff. But when you add two to the dimension of a square, not only are you adding the 3 dimensional amount of stuff, but you are also adding 4 dimensional amount of stuff now as well, and so on.


I posted one answer already, in which I provide my understanding of what the teacher is actually asking for.

That said, I also want to provide a weird take on this problem.

The pluses pile up faster.

  1. an integer bigger than zero is made of a pile of 1's and pluses.

$3 = 1 + 1 + 1$

$5 = 1 + 1 + 1 + 1 + 1$

$6 = 1 + 1 + 1 + 1 + 1 + 1$

  1. Adding two numbers together is just bringing all the pluses together.

$6 + 2 = (1 + 1 + 1 + 1 + 1 + 1) + (1 + 1)$

$6 + 3 = (1 + 1 + 1 + 1 + 1 + 1) + (1 + 1 + 1) $

$6 + 4 = (1 + 1 + 1 + 1 + 1 + 1) + (1 + 1 + 1 + 1) $

  1. Multiplication is repeated addition.

$2 \times 3$ means "add together two stacks of three."

$3 \times 3$ means "add together three stacks of three."

$1 \times 3 = (1 + 1 + 1)$

$4 \times 3 = (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1)$

$5 \times 3 = (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1)$

$6 \times 3 = (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1)$

The pluses pile up faster when you multiply than when you add things because you are adding them over and over.

  1. Exponentiation is repeated multiplication.

$3 ^ 4$ means is the same as $3 \times 3 \times 3 \times 3$, which means "add three stacks of three, then add together three stacks of that resulting one, then add together three stacks of THAT resulting one."

$3 ^ 1 = (1 + 1 + 1)$

$3 ^ 2 = (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1)$

$3 ^ 3 = (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1) + (1 + 1 + 1)$

The pluses pile up faster when you exponentiate things than when you multiply things because you are multiplying then over and over.

  1. A quadratic expression is just one multiplication.

$a ^ 2 = a \times a$

$b ^ 2 = b \times b$

$c ^ 2 = c \times c$

The pluses pile up in same speed as multiplication does.

  1. An exponential expression is many multiplications.

$a ^ 2 = a \times a$

$a ^ 3 = a \times a \times a$

$a ^ 4 = a \times a \times a \times a$

The pluses pile up in the same speed as exponentiation does, so an exponential expression piles up pluses faster than a quadratic expression.

  1. The speed in which an expression gets larger is equal to the number of new pluses it is getting at every step.

Multiplication gets more pluses each step than addition, as it is addition done over and over. Exponentiation gets more pluses each step than multiplication, as it is multiplication done over and over.

  1. The other numbers around don't change the speed, so they don't matter to see how fast things are growing.


$10 + 2 ^ 2 = (1+1+1+1+1+1+1+1+1+1) + (1+1) + (1+1) $

$10 + 3 ^ 2 = (1+1+1+1+1+1+1+1+1+1) + (1+1+1) + (1+1+1) + (1+1+1) $

$10 + 4 ^ 2 = (1+1+1+1+1+1+1+1+1+1) + (1+1+1+1) + (1+1+1+1) + (1+1+1+1) + (1+1+1+1) $

$10 + 5 ^ 2 = (1+1+1+1+1+1+1+1+1+1) + (1+1+1+1+1) + (1+1+1+1+1) + (1+1+1+1+1) + (1+1+1+1+1) + (1+1+1+1+1)$


$4 + 2 ^ 2 = (1+1+1+1) + (1+1) + (1+1) $

$4 + 2 ^ 3 = (1+1+1+1) + (1+1) + (1+1) + (1+1) + (1+1) $

$4 + 2 ^ 4 = (1+1+1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) $

$4 + 2 ^ 5 = (1+1+1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1) + (1+1)$

It's getting pluses way faster!

The number of pluses those extra numbers add are fixed, so they don't influence the speed of which the numbers grow. It may take some time, but eventually the exponential one will be getting pluses faster than the quadratic one.


Perhaps this is above the level of a 7th-grader, but a series expansion gives:

$$e^x > 1+x+\frac{x^2}{2}+\frac{x^3}{6}>\frac{x^3}{6}$$

and thus the cubic (and higher) terms will always dominate a quadratic at large $x$.

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    $\begingroup$ Writing any finite truncation with $\approx$ in the regime of large $x$ is bad practice all around, since what you neglected dominates over what you kept. The important thing in this approach (which is far above the level here anyway) is that $e^x$ is bigger than its finite truncations for positive $x$. $\endgroup$
    – Ian
    Feb 27, 2021 at 20:09
  • $\begingroup$ That's what I meant by "and higher." Right? $\endgroup$ Feb 27, 2021 at 20:38
  • $\begingroup$ Somehow this way of writing things makes it sound like the last term is "just a correction" but actually it's just the opposite in the regime we're talking about. $\endgroup$
    – Ian
    Feb 27, 2021 at 20:52
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    $\begingroup$ How would you explain/convince a 7th grader that $\exp(x) > 1 + x + \frac{x^2}{2} + \frac{x^3}{3}$? This is an interesting idea, but it feels like it might end up seeming like magic to a 7th grader. $\endgroup$ Feb 28, 2021 at 16:13
  • $\begingroup$ Yeah... I admit... that might be "magic" to a 7th grader. $\endgroup$ Feb 28, 2021 at 21:27

You might let $x$ get to infinity through only a thin sequence. If you want to compare $1.01^x$ with $100x^2$, suppose $x = 1.01^n$ and let $n$ go to infinity. Now you're comparing

$$100\cdot 1.01^{2n} \mbox{ with } (1.01)^{1.01^n}.$$

Now it's a matter of which exponent $2n$ or $1.01^n$ get bigger; does an exponential grow faster than a linear function? If you have to argue yes, you can repeat the above, and reduce to whether an exponential grows faster than a constant.


By shifting and rescaling, you can turn any comparison of a quadratic vs. exponential into a standard one. Say you are trying to determine which of $f(x) = \alpha(x-\beta)^2$ and $g(x) = ab^x$ grows faster. By shifting $x\rightarrow x+\beta$ you can turn this into a comparison of $f_{1}(x) = \alpha x^2$ and $g_{1}(x) = a' b^x$. Then by rescaling $x\rightarrow c x$ where $b^c = 2$, you turn the comparison into $f_2(x) = \alpha' x^2$ and $g_2(x) = a'' 2^x$.

So as long as you can convince your son that $2^x$ grows faster than $a x^2$, all other exponentials must also grow faster than all other quadratics.


We can show a more general result: a (growing) exponential grows faster than any power. Eventually,


Indeed, taking the logarithm, this is equivalent to

$$x\log a>b\log x$$


$$\frac x{\log x}>\frac b{\log a}.$$

The LHS is an unbounded function and the RHS a constant. (For $x>e$,

$$\frac{ex}{\log ex}=e\frac{x}{\log x+1}=e\frac{\log x}{\log x+1}\frac{x}{\log x}\ge\frac e2\frac{x}{\log x}.)$$

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