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I’m currently in my senior year of high school and we just started on the topic of logs, when doing textbook work I encountered a problem and I am confused on where I’m going wrong. Could any body help?

$$2^x+1 = 3^x-1 \implies x\log2 + \log2 = x\log3 - \log3$$ After this I don’t know where to go.

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  • $\begingroup$ Hi and welcome to math.SE. Please use MathJax formatting to improve readability and increase your chances to get meaningful answers. $\endgroup$ – francescop21 Oct 25 '18 at 15:52
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    $\begingroup$ After that is just a linear equation. You should be able to find $x$. $\endgroup$ – francescop21 Oct 25 '18 at 15:53
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Remember, the values $\log 2$ and $\log 3$ are just real numbers, so you're now trying to solve the equation $ax + a = bx - b$, where $a=\log 2$ and $b = \log 3$.

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It is $$x(\ln(2)-\ln(3))=-\ln(2)-\ln(3)$$

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HINT

$$\ldots \iff x\log3-x\log2 =\log3+\log2 $$

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Taking the log of the LHS and RHS is \begin{align*} \log(2^{x + 1}) &= \log(3^{x-1})\\ (x+1)\cdot\log(2) &= (x-1)\cdot\log(3)\\ x\log(2) + \log(2) &= x\log(3) - \log(3)\\ \Rightarrow x &= -\frac{\log(3) + \log(2)}{\log(2) - \log(3)} \approx 42.78468556 \end{align*}

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