 # Struggling to finish an exponential equation I have this exponential equation I tried solving but I'm stuck at the end. The problem is this: 4^x-3^(x-1/2)=3^(x+1/2)-2^(2x-1) blogswput 2022-09-11 Answered
Struggling to finish an exponential equation
I have this exponential equation I tried solving but I'm stuck at the end.
The problem is this:
${4}^{x}-{3}^{x-\frac{1}{2}}={3}^{x+\frac{1}{2}}-{2}^{2x-1}$
And my approach is this:
${2}^{2x}+{2}^{2x-1}={3}^{x+\frac{1}{2}}+{3}^{x-\frac{1}{2}}$
${2}^{2x}\left(1+{2}^{-1}\right)={3}^{x}\left({3}^{\frac{1}{2}}+{3}^{-\frac{1}{2}}\right)$
${2}^{2x}\ast \frac{3}{2}={3}^{x}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)$
${2}^{2x}\ast \frac{3}{2}={3}^{x}\left(\frac{{\sqrt{3}}^{2}+1}{\sqrt{3}}\right)$
${2}^{2x}\ast \frac{3}{2}={3}^{x}\ast \frac{4\sqrt{3}}{3}$--> here I have rationalized the fraction
$\frac{{2}^{2x}}{{3}^{x}}=\frac{4\sqrt{3}}{3}\ast \frac{2}{3}$
${\left(\frac{{2}^{2}}{3}\right)}^{x}=\frac{8\sqrt{3}}{9}$
Now I hope I didn't make an error. How should I proceed from this point? Is there another way around this problem?
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I think you did not make a mistake.
Your last equation it's
${\left(\frac{2}{\sqrt{3}}\right)}^{2x}={\left(\frac{2}{\sqrt{3}}\right)}^{3}$
because
$\frac{8\sqrt{3}}{9}=\frac{8\sqrt{3}}{\sqrt{8}1}=\frac{8}{\sqrt{27}}=\frac{{2}^{3}}{\sqrt{{3}^{3}}}=\frac{{2}^{3}}{{\left(\sqrt{3}\right)}^{3}}={\left(\frac{2}{\sqrt{3}}\right)}^{3}.$

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