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.

Answer 1

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 $$x=\frac{-8-\log_64}{1-\log_64}$$

Answer 2

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:

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

Answer 3

1. Remove the logs of exponents

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

2. Rewrite bases

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

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

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

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

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

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

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