# Exercise ID327

Algebra → Logarithm → Logarithm equations
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Evaluate the following expression:

${2}^{2{\mathrm{log}}_{10}\left(4x-1\right)}-{7}^{{\mathrm{log}}_{10}\left(4x\right)}={7}^{{\mathrm{log}}_{10}\left(4x-1\right)}-3×{4}^{{\mathrm{log}}_{10}\left(4x\right)}\phantom{\rule{0ex}{0ex}}$

$t={\mathrm{log}}_{10}\left(4x\right)\phantom{\rule{0ex}{0ex}}$ ${2}^{2t-1}-{7}^{t}={7}^{t-1}-3×{4}^{t}\phantom{\rule{0ex}{0ex}}$ ${2}^{2t}×{2}^{-1}-{7}^{t}={7}^{t}×{7}^{-1}-3×{4}^{t}\phantom{\rule{0ex}{0ex}}$ ${4}^{t}×\frac{1}{2}-{7}^{t}={7}^{t}×\frac{1}{7}-3×{4}^{t}\phantom{\rule{0ex}{0ex}}$ $\frac{1}{2}-\frac{{7}^{t}}{{4}^{t}}=\frac{{7}^{t}}{7×{4}^{t}}-3\phantom{\rule{0ex}{0ex}}$ $\frac{1}{2}-{\left(\frac{7}{4}\right)}^{t}=\frac{1}{7}×{\left(\frac{7}{4}\right)}^{t}-3\phantom{\rule{0ex}{0ex}}$ $-{\left(\frac{7}{4}\right)}^{t}-\frac{1}{7}×{\left(\frac{7}{4}\right)}^{t}=-3-\frac{1}{2}\phantom{\rule{0ex}{0ex}}$ ${\left(\frac{7}{4}\right)}^{t}=\frac{49}{16}\phantom{\rule{0ex}{0ex}}$ ${\left(\frac{7}{4}\right)}^{t}={\left(\frac{7}{4}\right)}^{2}\phantom{\rule{0ex}{0ex}}$ $t=2\phantom{\rule{0ex}{0ex}}$ ${\mathrm{log}}_{10}\left(4x\right)=t⇒{\mathrm{log}}_{10}\left(4x\right)=2\phantom{\rule{0ex}{0ex}}$ ${\mathrm{log}}_{10}\left(4x\right)={\mathrm{log}}_{10}\left({10}^{2}\right)\phantom{\rule{0ex}{0ex}}$ $4x=100\phantom{\rule{0ex}{0ex}}$ $x=25\phantom{\rule{0ex}{0ex}}$
$x=25\phantom{\rule{0ex}{0ex}}$