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How can this equation be solved: $2^{x+1}=3^{2x-1}$

I can get this far:

$2^x(2^1) = \frac{3^{2x}}{3^1}$

$2(2^x) = \frac{1}{3}(3^{2x})$

I know I need to log both sides to remove the $x$'s but not sure how it would work with to $2$ and $1/3$.

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    Note: $3=2^{\log_2(3)}$. – 2017-01-16
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    Yes, you're on the right track. Use $\log(ab) = \log(a) + \log(b)$ to proceed further or otherwise take log in the given equation itself. – 2017-01-16
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    @projectilemotion Please undelete your answer. I was in the middle of answering your question in a comment when you deleted your answer. Your answer is/was fine and correct. Perhaps my first comment to you should have been framed differently. It is better to have a range of correct answers which may vary in approach. Please undelete your very correct answer, so I can upvote it? – 2017-01-16

5 Answers 5

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We have: $$2^{x+1}=3^{2x-1}$$ We take the logarithm base $2$ on both sides: $$\log_2{(2^{x+1})}=\log_2{(3^{2x-1})}$$ We use the identity $\log(a^b)=b\log(a)$: $$x+1=(2x-1)\cdot\log_2{3}$$ Expand: $$x+1=(2\log_2{3})x-\log_2{3}$$ Gather terms: $$x-(2\log_2{3})x=-\log_2{3}-1$$ Factorize: $$(1-2\log_2{3})x=-(\log_2{3}+1)$$ $$(2\log_2{3}-1)x=\log_2{3}+1$$ Giving us the solution: $$\boxed{x=\frac{\log_2{3}+1}{2\log_2{3}-1}}$$

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$$(x+1)\log2=(2x-1)\log3$$

Now $\log(ab/c)=\log a+\log b-\log c$

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Take the logarithm of base $2$

$(x+1)log(2)=(2x-1)log(3)$

Simplify

$x+1=2log(3)x-log(3)$

The rest is easy

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$$\log_{2}2^{x+1}=\log_{2}3^{2x-1} \rightarrow x+1=(2x-1)\log_2 3$$

$$x(1-2\log_2 3)=-1-\log_2 3 \rightarrow x=\frac{1+\log_2 3 }{2\log_2 3-1}$$

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Full solution:

Start with the $(x+1)\log 2=(2x-1)\log 3$ (as stated by lab bhattacharjee).

Then $\frac{x+1}{2x-1}\frac{\log 2}{\log 3}=1$.

So $\frac{x+1}{2x-1}=\frac{\log_33}{\log_32}=\log_23$.

So $x+1=2x\log_23-\log_23.$

This gives $(2\log_23-1)x=1+\log_23$ so $x=\frac{\log_26}{\log_2\frac{9}{2}}=\log_{\frac{9}{2}}6.$

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    $1+\log_2 3 = \log_2 6$, not $\log_2 3$. – 2017-01-16
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    fixed it, thx :3 – 2017-01-16