Calculating Sums Of Maths

Question

I was wondering whether I got the correct answer. I'm supposed to solve by logs and get the exact value of x using my calculator. My question:

$9^{x -5} = 2^{x - 8}$

Answer 1

We have:

$$6^{x-2}=4^{x-1}\to(2^{x-2})(3^{x-2})=2^{2x-2}$$ $$\to3^{x-2}=2^{x}$$

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We take logs so that: $$x=\log_{2}{3^{x-2}}$$ $$\to x=(x-2)\log_2{3}$$

Then use an iteration:

$$x_{n+1}=(x_n-2)\log_2{3}$$

Plugging in values proves this iterate diverges.

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We can also take logs for: $$(x-2)=\log_3{2^x}$$ $$\to x=x\log_3{2}+2$$ and use the iteration: $$x_{n+1}=x_n\log_3{2}+2, x_0=1$$ We get $$\lim_{n\to\infty}x_n\approx 5.419022583...$$ and so this is a solution. The divergence of the first iterate shows that this is the only solution.

Answer 2

So you will get $$(x-2)\ln(6)=(x-1)\ln(4)$$ you will get $$x(\ln(6)-\ln(4))=2\ln(6)-\ln(4)$$