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$x$ should equal to.. I don't know how to get to the answer.

$x= \log_{10} (2\pm \sqrt{3})$

Thank you.

Similarly: $7^x+7^{-x}=4$. The answer is the same.

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    Presumably the answer to the second question is $\log_{7} (2\pm \sqrt{3})$ – 2017-01-22

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Hint : Use $a^{-x}=\frac{1}{a^x}$ and substitute $t=a^x$ to get a quadratic equation

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    ill give it a go thank you – 2017-01-22
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    #Peter - could you show me how you work it out, i cant get to that answer. Thanks – 2017-01-23
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    First equation : Set $t=10^x$. You get $t+\frac{1}{t}=4$ implying $t^2+1=4t$, hence $t^2-4t+1=0$. Solving the quadratic equation gives $t=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt{3}$. Now you have $x=\log_{10}(t)$ – 2017-01-24
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    The second equation is analogue, but here we have $t=7^x$, hence $x=\log_7(t)$ – 2017-01-24