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I understand how to solve a problem such as $6^{2x+3}=11$ by using natural log, but the question thats tripping me up is $3^{1-x}=7^x$. Mathway and Wolfram Alpha tells me what the answer is, but I cant seem to reproduce that answer. Any help?

The Question im being asked is : Solve the exponential equation. Write the exact answer with natural logarithms and then approximate the result correct to three decimal places. $$3^{1-x}=7^x$$

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Taking natural log from both side gives: $$\ln{3}(1-x) = x\ln{7}\\ x(\ln{3} + \ln{7}) = \ln{3}\\ x = \frac{\ln{3}}{\ln{3} + \ln{7}}=\frac{\ln{3}}{\ln21}$$ Then use a calculator to get your desired answer as 3 decimal places.

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Notice, $$3^{1-x}=7^x$$ taking log both the sides, $$\ln 3^{1-x}=\ln 7^x$$ $$(1-x)\ln 3=x\ln 7$$ $$x(\ln 3+\ln 7)=\ln 3$$ $$x\ln(3\times 7)=\ln 3$$ $$x=\color{red}{\frac{\ln 3}{\ln21}}$$

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