# Find the value of x such that

Find the value of $x$ such that ${2}^{x}=10$
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Immanuel Glenn
We know,
${\mathrm{log}}_{10}10=1$
But
${\mathrm{log}}_{10}10={\mathrm{log}}_{10}\left(2\cdot 5\right)={\mathrm{log}}_{10}2+{\mathrm{log}}_{10}5$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{10}2=1-{\mathrm{log}}_{10}5=1-0.7=0.3$
Now taking logarithm on the given equation
$x{\mathrm{log}}_{10}2={\mathrm{log}}_{10}10=1$
as ${\mathrm{log}}_{a}\left({b}^{m}\right)=m{\mathrm{log}}_{a}b$

delirija7z
${2}^{x}=10/\cdot {\mathrm{log}}_{10}$
${\mathrm{log}}_{10}{2}^{x}={\mathrm{log}}_{10}10$
$x{\mathrm{log}}_{10}2=1$
$x=\frac{1}{{\mathrm{log}}_{10}2}=\frac{1}{{\mathrm{log}}_{10}\frac{10}{5}}=\frac{1}{{\mathrm{log}}_{10}10-{\mathrm{log}}_{10}5}=\frac{1}{1-{\mathrm{log}}_{10}5}=\frac{1}{1-0.7}=\frac{1}{0.3}=\frac{1}{\frac{3}{10}}=\frac{10}{3}$
$x=\frac{10}{3}$

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