Question

$$\frac{1^{5}}{5}\times\frac{1}{5}^{19}=\frac{1^{8x}}{5}$$

Answer

x=INF

Solution


Simplify  \({1}^{5}\)  to  \(1\).
\[\frac{1}{5}\times \frac{1}{{5}^{19}}=\frac{{1}^{8x}}{5}\]
Simplify  \({5}^{19}\)  to  \(1.907349\times {10}^{13}\).
\[\frac{1}{5}\times \frac{1}{1.907349\times {10}^{13}}=\frac{{1}^{8x}}{5}\]
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{1\times 1}{5\times 1.907349\times {10}^{13}}=\frac{{1}^{8x}}{5}\]
Simplify  \(1\times 1\)  to  \(1\).
\[\frac{1}{5\times 1.907349\times {10}^{13}}=\frac{{1}^{8x}}{5}\]
Simplify  \(5\times 1.907349\)  to  \(9.536743\).
\[\frac{1}{9.536743\times {10}^{13}}=\frac{{1}^{8x}}{5}\]
Multiply both sides by \(5\).
\[\frac{1}{9.536743\times {10}^{13}}\times 5={1}^{8x}\]
Simplify  \(\frac{1}{9.536743\times {10}^{13}}\times 5\)  to  \(\frac{5}{9.536743\times {10}^{13}}\).
\[\frac{5}{9.536743\times {10}^{13}}={1}^{8x}\]
Use Definition of Common Logarithm: \({b}^{a}=x\) if and only if \(log_b(x)=a\).
\[\log_{1}{(\frac{5}{9.536743\times {10}^{13}})}=8x\]
Use Change of Base Rule: \(\log_{b}{x}=\frac{\log_{a}{x}}{\log_{a}{b}}\).
\[\frac{\log{(\frac{5}{9.536743\times {10}^{13}})}}{\log{1}}=8x\]
Use Rule of One: \(\log{1}=0\).
\[\frac{\log{(\frac{5}{9.536743\times {10}^{13}})}}{0}=8x\]
Simplify  \(\frac{\log{(\frac{5}{9.536743\times {10}^{13}})}}{0}\)  to  \(\infty \).
\[\infty =8x\]
Divide both sides by \(8\).
\[\frac{\infty }{8}=x\]
Simplify  \(\frac{\infty }{8}\)  to  \(\infty \).
\[\infty =x\]
Switch sides.
\[x=\infty \]