Solve $6^{x+8} = 4^{x-1}$

Question

I tried doing $log_6\left(6^{x+8}\right) = log_6{4^{x-1}}$

I got stuck, and I don't think that was the right route.

For all variations of your problem of the kind (e.g., with $x+8$ replaced by $x+3$) the solution can never be in whole numbers. Because LHS would be divisible by 3 and the RHS would not be. — P Vanchinathan, May 20 '15 at 00:29

score 6 · Answer 1 · answered May 20 '15 at 00:21

6

The OP's start was fine. We rewrite the left side as $x+8$, and pull the exponent out of the right side, to get $$x+8=(x-1)\log_64$$ We expand the right side to get $$x+8=x(\log_64)-\log_64$$ then subtract $8+x\log_64$ from both sides, to get $$x-x\log_64=-8-\log_64$$ We now factor out $x$ to get $$x(1-\log_64)=-8-\log_64$$ Lastly, we divide by $1-\log_64$ to get $$x=\frac{-8-\log_64}{1-\log_64}$$

answered May 20 '15 at 00:21

vadim123

82,796

Can't I use change of base here to get a number? – Bassinator May 20 '15 at 01:05
This is a number, albeit a somewhat messy one. It is possible to simplify the expression. – vadim123 May 20 '15 at 02:21

Eoin · Answer 2 · 2015-05-20T00:34:43.337

3

It doesn't necessarily matter which base you are in, you can solve this as follows $$6^{x+8}=4^{x-1}\iff \log(6^{x+8})=\log(4^{x-1})$$ then, by rules of the logarithm, $$(x+8)\log 6=(x-1)\log 4$$ now, by the distributive property for real numbers, $$(x+8)\log 6=(x-1)\log 4 \iff x\log 6 + 8\log 6 = x\log 4 -\log4$$ and so

$$x\log6-x\log 4=-8\log 6 - \log4$$ again by the distributive property $$x(\log6-\log4)=-8\log 6 - \log4$$ and since $(\log 6 - \log 4)\neq 0$, we can divide both sides by it to achieve:

$$x=\frac{-8\log 6 - \log 4}{\log6-\log4}$$

edited May 20 '15 at 00:34

answered May 20 '15 at 00:24

Eoin

5,809

1

Note that in base $6$ we get vadim123's answer. I posted this because (partially I had almost finished it when another answer was posted) and I want to emphasize the interelation between exponents and the logarithm. The base of the logarithm is not so important as to what it is meant for. – Eoin May 20 '15 at 00:25

A.J. Uppal · Answer 3 · 2015-05-20T03:25:35.713

0

Remove the logs of exponents

$$(6x+8)*\log_6(6)=(x-1)*\log_6(4)$$

Rewrite bases

$$(6x+8)*\frac{\log(6)}{\log(6)}=(x-1)*\frac{\log(4)}{\log(6)}$$

Since $\frac{\log(6)}{\log(6)} = 1$,

$$(6x+8)=(x-1)*\frac{\log(4)}{\log(6)}$$

Thus, dividing $\frac{(6x+8)}{(x-1)}$:

$$\frac{(6x+8)}{(x-1)}=\frac{\log(4)}{\log(6)}$$

Then, after further solving the equation, we get the value of $x$:

$$x=-\frac{\log(2592)}{\log(108)}$$

edited May 20 '15 at 03:25

answered May 20 '15 at 00:30

A.J. Uppal

760

The original equation was $6^{x + 8} = 4^{x - 1}$. If you take base $6$ logarithms of both sides, you should obtain $(x + 8)\log_6 6 = (x - 1)\log_6 4$. – N. F. Taussig May 20 '15 at 09:21

Solve $6^{x+8} = 4^{x-1}$

3 Answers3