This is the question : enter image description here

Find all natural numbers $n$ so that $6^{n+2}≤7^{n-1}$. Prove your claim using induction.

I'm not sure how to proceed with this. I have solve the equation $6^{n+2}\leq 7^{n-1}$ to get $$n > 35.8$$ so for $n \geq 36$ onwards the inequality holds true. So does that mean I start with base case $36$ and then continue from there or do I need to show the working for solving the equation to get $n\geq 36$ before beginning to use induction.

And for the induction step after we assume claim holds true for $n = k$, I am not sure how to do the induction step where we have to use the assumption for $n=k$ to prove $n=k+1$.

  • 2
    $\begingroup$ Yes, of course the base case then is $n=36$ (for smaller $n$ it is false). $\endgroup$ Commented Nov 9, 2022 at 17:58
  • 2
    $\begingroup$ I don't know why you'd want to prove it by induction. The argument you used to find $36,$ presumably using somethiing like $\log_{7/6}(6^2\cdot 7),$ is enough to show it for all $n\geq 36.$ And induction is going to be hard to prove the inequality is not true for $n<36.$ $\endgroup$ Commented Nov 9, 2022 at 18:02
  • 2
    $\begingroup$ @Thomas Andrews, it may just be that the author has to use induction (perhaps for a requirement for an assignment). In which case, for insight on how to go about proof by induction, see people.vcu.edu/~rhammack/BookOfProof/Main.pdf chapter 10. $\endgroup$ Commented Nov 9, 2022 at 18:03
  • $\begingroup$ I mean, the equation is equivalent to: $$6^2\cdot 7\leq\left(\frac76\right)^n.$$ So if it is true for $n=36,$ it is clearly true for all bigger $n.$ I guess you can prove that by induction. So you can prove it is true for $n=36$ and not true for $n=35,$ then deduce by induction 'both ways' that it is true exactly for when $n\geq 36.$ $\endgroup$ Commented Nov 9, 2022 at 18:08
  • 1
    $\begingroup$ This seems like a backwards question. It seems you have to know enough about exponents to solve but if you know enough about exponents to solve you don't need induction to prove. None the less the induction is simple: If $6^{k+2} \le 7^{k-1}$ then $6^{(k+1)+2}=6^{k+2}\cdot 6 \le 7^{k-1}\cdot 6<7^{k-1}\cdot 7 =7^{(k+1)-1}$. So if your base case is $36$ you ahve proven all $n\ge 36$ are solutions. But you must also prove $35$ is not a solution. (Then by the contrapositive on $k \le 35$ can be a solution because if $k \le 35$ is a solution you have proven by induction $35$ is a solution) $\endgroup$
    – fleablood
    Commented Nov 9, 2022 at 18:16

3 Answers 3


The logarithmic inequality shows that, there exist the smallest $n=k,\,k\in\mathbb Z^{+}$, such that $7^{k-1}≥6^{k+2}$ holds.

Let $7^{k-1}-6^{k+2}=a,\,a>0$. Then you can construct an induction hypothesis.

For $n=k$, then the statement is correct. Then for $k+1$, we have

$$ \begin{align} 7^k-6^{k+3}&=7\left(a+6^{k+2}\right)-6^{k+3}\\ &=7a+7\cdot 6^{k+2}-6^{k+3}\\ &≥7a+6^{k+3}-6^{k+3}\\ &=7a>0.\end{align} $$

This completes the proof.


I think you are supposed to use induction to prove that once you found the first one you've found them all.

That is: Induction step: If $k$ is a solution so that $6^{k+2} \le 7^{k-1}$ then $6^{(k+1)+2} = 6^{k+2}\cdot 6 \le 7^{k-1} \cdot 6 < 7^{k-1}\cdot 7 = 7^k$.

Thus if $k$ is the smallest natural number that is a solution (if such a solution exist) then you have proven that the solution set is all natural numbers greater than or equal to $k$.

So is seems your base case is to show that $6^{36+2} \le 7^{36-1}$ and ... I'm really not sure what you are expected to know to prove that.

But you must also show that $6^{35+2} > 7^{35-1}$ and that $35$ is not a solution. By contrapositive that would mean that for any $k < 36$ is not a solution. (If $k\le 35 < 36$ is a solution then by induction $35$ would be a solution and it is not).

I'm really not sure how the text intends for you to do that as ... well if you can do it you can prove you have them all without resorting to induction. (I.E use rules of exponents and logarithms.)


Alternative approach:

Once the base case $~k \in \Bbb{Z^+}~$ is established so that $~6^{(k+2)} \leq 7^{(k-1)},~$ induction may be used to complete the problem with virtually no Math.

Make the induction hypothesis that for a specific $N \in \Bbb{Z^+}:$

$$6^{(N+2)} \leq 7^{(N-1)}. \tag1 $$

You are then supposed to use the assumption represented by (1) above to show that the assertion must then also be true for $(N+1).$ This means that you are supposed to use the assumption represented by (1) above to show that

$$6^{([N+1]+2)} \leq 7^{([N+1]-1)}. \tag2 $$

However, the assertion in (2) above is an immediate consequence of the assertion in (1) above, because:

  • The LHS of (2) above represents $6 \times$ the LHS of (1) above.
  • The RHS of (2) above represents $7 \times$ the RHS of (1) above.
  • $6 < 7.$
  • 1
    $\begingroup$ Just as a rule, there is no reason to put parentheses around exponents like this. $6^{k+2}$ is already unambiguous. $\endgroup$ Commented Nov 9, 2022 at 18:46
  • $\begingroup$ @ThomasAndrews I am harmlessly anal. $\endgroup$ Commented Nov 9, 2022 at 18:53
  • $\begingroup$ But it isn't harmless - it adds a slight bit more work to read, and it makes the reader wonder if you have written what you meant to write. As a rule, try to adjust your anal retentive tendencies towards clarity of communication, rather than some internal needs. (I am anal, too, otherwise I wouldn't have commented.) $\endgroup$ Commented Nov 9, 2022 at 18:58
  • $\begingroup$ @ThomasAndrews I was trying to be polite. You are looking at the writing through sophisticated eyes, which will generally be able to pierce the parentheses to discern intent. My real target audience is the high school senior or college freshman. Note how the use of parenthesis in (1) clarifies (to the freshman) the expression in (2). I have had this disagreement with others. It is difficult to take such a viewpoint as yours seriously, because (in my opinion) over $50\%$ of the answers given to (high quality - deserving) postings substitute brevity for clarity. ...see next comment $\endgroup$ Commented Nov 9, 2022 at 20:28
  • $\begingroup$ That is, most of the answers given are nowhere near long-winded enough to make it absolutely clear, at a glance, to the freshman, that the answer is valid and accurate. Perhaps, assuming that you don't insist on having the last word, that we should just call it a day, and agree to disagree. $\endgroup$ Commented Nov 9, 2022 at 20:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .