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E.g. I have the question:

$$2^{4x+1} = 128$$

I solved this by knowing that $128 = 2^7$ and therefore $x$ must equal $1.5$.

However, is there a way of solving this without knowing that $128 = 2^7$?

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    $\begingroup$ $$128 = 2^{7}$$ $\endgroup$ – Shreyas S Dec 14 '16 at 17:38
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    $\begingroup$ $x=2$ doesn't work. There is no need to "know" that $128=2^7$...if you wanted to solve $2^y=z$ you could just take $\log_2$ of both sides to get $y=\log_2 z$. $\endgroup$ – lulu Dec 14 '16 at 17:41
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    $\begingroup$ Well, now that you've added "without knowing that $128 = 2^7$"... What exactly do you mean by that??? If you take $\log_2$ on $128$, and you get $7$, then does that mean that you "know that $128 = 2^7$? $\endgroup$ – barak manos Dec 14 '16 at 17:43
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    $\begingroup$ The point is that there should be a way of answering the question without knowing already that 128 = 2^7. $\endgroup$ – top_systems_programmer Dec 14 '16 at 17:48
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    $\begingroup$ If you didn't already know that $128=2^7$, you would begin by finding the prime factorization of $128$. $\endgroup$ – Jack M Dec 14 '16 at 23:28
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... and without logarithms or knowing any powers of $2$ other than the most trivial one ... \begin{align*} 2^{4x+1} &= 128 \\ 2^{4x+1-1} = 2^{4x} &= 128/2 = 64 \\ 2^{4x-1} &= 32 \\ 2^{4x-2} &= 16 \\ 2^{4x-3} &= 8 \\ 2^{4x-4} &= 4 \\ 2^{4x-5} &= 2 \\ 2^{4x-6} &= 1 = 2^0 \text{,} \\ \end{align*} so $4x-6 = 0$ and $x = 6/4 = 3/2$.

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    $\begingroup$ This is actually the real solution. Any solution involving log mean you must know that 128 is 2^7 so this should be marked as answer $\endgroup$ – Thaina Dec 15 '16 at 11:42
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    $\begingroup$ @Thaina: See my comments (and the comment-thread in general) on the question itself. In addition, for this method to work, $128$ has to be a perfect power of $2$, and whoever deploys this method needs to know it. So in terms of "knowing", there is not much difference between using $\log_2$ and using this method. $\endgroup$ – barak manos Dec 15 '16 at 11:43
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    $\begingroup$ I think he should've stopped at $2^{4x-5}$, as establishing $2^0=1$ may not be as obvious as $2^1=2$. $\endgroup$ – Simply Beautiful Art Dec 16 '16 at 0:55
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    $\begingroup$ @Thaina Eh, no, replace 128 in the question by any arbitrary number, say 185, and the log based solutions still work. $\endgroup$ – Masked Man Dec 16 '16 at 8:32
  • $\begingroup$ @Thaina - the solutions using log don't require knowing that 128 is 2^7, in fact it's the opposite, it provides a general solution $\endgroup$ – Cato Dec 20 '16 at 18:09
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$2^{4x+1}=128\iff$

$\log_22^{4x+1}=\log_2128\iff$

${4x+1}=7\iff$

${4x}=6\iff$

${x}=6/4$

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    $\begingroup$ This is just giving a different name to "knowing that $128=2^7$". $\endgroup$ – Jack M Dec 14 '16 at 23:27
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    $\begingroup$ @Jack M: Not if the step from second to third line has been done by actually computing the $\log_2$ of 128; of course after that you probably know it. $\endgroup$ – Curd Dec 15 '16 at 0:16
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    $\begingroup$ @JackM: See my comments (and the comment-thread in general) on the question itself. $\endgroup$ – barak manos Dec 15 '16 at 0:28
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    $\begingroup$ @JackM To put it another way - simply "knowing" that 128=2^7 is a piece of declarative information. Computing the logarithm of 128 is applying a functional algorithm to come to the same answer. If you know how to compute a logarithm you can "forget" that you ever knew any specific solutions. $\endgroup$ – J... Dec 15 '16 at 17:25
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$2^{4x+1}=128$

$\log 2^{4x+1}=\log 128$

$(4x + 1) \times \log 2 = \log 128$ - from properties of logs

$x = \frac{1}{4}(\frac {\log 128}{log (2)} - 1) = 3/2$

note that you can use any logarithm, log base 10 or 'ln' - or any other 'base' of logarithms you might have (with log10 and loge being the commonly found ones on calculators, spreadsheets etc ) you have to use your chosen type of log consistently of course

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    $\begingroup$ I think this answer is best beacuse it's the only one to point out that you do not need to use log-base-2. Given the OPs wording, this strikes me as a key insight towards what the OP wanted. $\endgroup$ – Cort Ammon Dec 15 '16 at 1:18
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To solve your equation, take the base-2 logarithm of both sides:

$$ \log_2 2^{4x+1} = \log_2 128 $$ $$ 4x+1 = 7 $$ $$ x = 1.5$$

However, is there a way of solving this without knowing that $128=2^7$?

Well, this piece of information is equivalent to “knowing that $\log_2 128 = 7$”, so no.

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    $\begingroup$ calculators don't often have log base 2, you can just use any type of log, log 10 or ln(base e) both work. lol, when he says 'know' he means 'be able to spot', not work it out with a calculator $\endgroup$ – Cato Dec 14 '16 at 17:54
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    $\begingroup$ @Cato, but there's always the base change formula, y'know. $\endgroup$ – J. M. is a poor mathematician Dec 15 '16 at 13:58
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It's very easy. Let's assume we have

$$k_1 ^ {k_2 * x + k_3} = k_4 $$

Where all the $k$'s are knowns, and the x is unknown.

We begin by removing the $k_3$ from the result ($k_4$) by dividing it by $k_1 ^ {k_3}$. The result thus obtained should be equal to:

$$k_1 ^ {k_2 * x}$$

As you can see we no longer have to deal with the $k_3$. Next we raise the result above to the power of $\frac{1}{k_2}$. That is, we find out what number we have to raise to the power of $k_2$ to reach at the result $k_1 ^ {k_2 * x}$. The answer should be equal to:

$$k_1 ^ x$$

Now finally from this result we can find out the value of $x$ by using a mathematical operation that lets us know, what number you have to raise a known other number to, to reach at a known result.

This operation is the logaritm. It's method of use is as follows: Let's assume $a ^ b = c$. The logarithm can be used to find out $b$, and it is then written as $\log_a c$.

So, to find out x in the result above, we use $\log_{k_1} result$ where $result$ is the value of the result obtained from previous operations.

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  • $\begingroup$ This is actually (the start of) the best answer, because it provides a step-by-step method for driving at the result of any problem of the sort posed by the asker. $\endgroup$ – S. G. Dec 15 '16 at 17:40
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$$2^{4x+1}=128\\\vphantom{\cfrac11}4x+1=7\\\vphantom{\cfrac11}4x=6\\\vphantom{\cfrac11}x=\frac64=\frac32$$

This is done as one would usually have:

$$2^y=128\iff y=\log_2(128)=7$$

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  • $\begingroup$ That's not what OP meant; see the comment-thread on the question. $\endgroup$ – barak manos Dec 14 '16 at 17:41
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    $\begingroup$ @barakmanos fixed. $\endgroup$ – Simply Beautiful Art Dec 14 '16 at 17:42
  • $\begingroup$ Well, the OP has just added "without knowing that $128 = 2^7$" (whatever that means)... $\endgroup$ – barak manos Dec 14 '16 at 17:44
  • $\begingroup$ @barakmanos ah the frustration... $\endgroup$ – Simply Beautiful Art Dec 14 '16 at 17:46
  • $\begingroup$ knowing that $128^7$ would mean to me that it is a fact that can be spotted by inspection, as opposed to $3^4.416508=128$ which isn't likely to be figured out without a calculation. $\endgroup$ – Cato Dec 14 '16 at 17:59
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This answer is based on that of user504882, and may be considered a heavily-edited (and extended) version thereof.

Let's assume we have

$$k_1 ^ {k_2 * x + k_3} = k_4, $$

where all the $k$s are known and the $x$ is unknown. As with almost all problems in elementary algebra, the goal is to isolate (i.e., solve for) $x$, and each step we take will be with that in mind.

We begin by removing the $k_3$ from the left-hand side of the equation by dividing it by $k_1 ^ {k_3}$. The equation thus obtained is:

$$k_1 ^ {k_2 * x} = \frac{k_4}{k_1 ^ {k_3}}.$$

As you can see, we no longer have to deal with the $k_3$ in the left-hand side. Next we further isolate $x$ on the left-hand side by raising both sides of the equation to the power of $\frac{1}{k_2}$. The resulting equation is:

$$k_1 ^ x = {(\frac{k_4}{k_1 ^ {k_3}})}^{\frac{1}{k_2}}.$$

Now finally we can isolate $x$ by using a mathematical operation that lets us invert exponentiation; i.e., it lets us know what number we have to raise a given number to the power of in order to reach at a known result.

This operation is the logarithm. Its definition is as follows: Let's assume $a ^ b = c$. The logarithm can be used to find out $b$, and it is then written as $$b = \log_a(c).$$

(Note that $a$ is called the base of $\log_a$, since it's the base of the exponential expression $a^b$.)

So, to isolate $x$ in the equation above, we use $\log_{k_1}$ to produce the equation $$x = \log_{k_1}({(\frac{k_4}{k_1 ^ {k_3}})}^{\frac{1}{k_2}}).$$

Note that logarithm rules can simplify this formula to

$$x = \frac{\log_{k_1}(k_4) - k_3}{k_2}.$$

As others have pointed out, however, when applied to the original problem, all of this fancy shuffling reduces to knowing that $128 = 2^7$ or, in the parlance of logarithms, $7 = \log_2(128).$ But what if we didn't have this special knowledge? What if the original equation were something like $54^{17x + 99} = 42$? How the heck could we reasonably solve that equation?

We do this by manipulating logarithms. It turns out that we've found two types of logarithms really useful, to the point that most standard calculators come equipped with them: the logarithms based in the number 10 and the well-known constant $e$. These both make sense, because our everyday number system uses 10s, and $e$ has some special properties that aren't relevant to this particular discussion. Regardless, $\log_e$ is used so often it's often called the natural logarithm, and the base is usually omitted (so that when you see $\log(c)$ it's understood to mean $\log_e(c)$).

So now, since we can easily work with logarithms in either of these bases (even if we have no idea what $e$ actually is or what it's otherwise used for), we just have to wonder whether or not there's a way for turning a logarithm based in an arbitrary number, say $a$, to a logarithm based in $e$ (or 10). And there is! Here's a formula for changing bases on a logarithm, from base $a$ to base $t$ (for arbitrary values of $a$ and $t$):

$$\log_{t}(c) = \frac{\log_a(c)}{\log_a(t)}.$$

(This formula isn't too difficult to prove, but it relies on some properties of logarithms that are also beyond the scope of this discussion, so we make take it as given.)

Augmenting our cooked-up formula with the change of base formula gets us

$$ x = \frac{\log_{k_1}(k_4) - k_3}{k_2} = \frac{\frac{\log(k_4)}{\log(k_1)} - k_3}{k_2}.$$

Now, that looks like a bit of intimidating alphabet soup, but it's not too difficult to replace all the $k$s with their respective numbers from the original problem. This yields $k_1 = 2$, $k_2 = 4$, $k_3 = 1$, and $k_4 = 128.$ Thus

$$ x = \frac{\frac{\log(128)}{\log(2)} - 1}{4} = \frac{7 - 1}{4} = \frac{6}{4} = 1.5.$$

It is left as an exercise for the reader to solve $54^{17x + 99} = 42$ for $x$, using this or any other method.

A question one can play around with: Is it possible to generalise our formula to quadratic polynomials? To general polynomials in the reals? To general complex polynomials? If you know, you may want to answer my question Name and Generalise This Linear Exponent Formula.

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I would go with the bisection method, because the next problem may not have nice integer solutions. For example, what if the right side of the equation is 138 instead of 128?

So start f(x) = 2^(4*x+1). Let L = 0 and U = 10. Verify that f(L) < 128 and also f(U) > 128.

Calculate m = (L+U)/2 and f(m). If f(m) < 128, let L=m and repeat. If f(m)>128, let U=m and repeat. If f(m)=0, let x=m and quit.

This process may never get you an exact solution, but L and U will approach each other and sandwich the solution close enough for all practical purposes.

I confess, this was inspired by my recent discovery of how Euler computed logarithms to 40 or more places.

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