Question

$$\sqrt { 2401 } = \sqrt { 7 ^ { x } } ,$$

Answer

$$x=log(7,(2401*If^2)/(e^2*t^2*h^2*n^2))$$

Solution


Since \(49\times 49=2401\), the square root of \(2401\) is \(49\).
\[If\times 49=\sqrt{{7}^{x}}then\]
Use this rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[If\times 49={7}^{x\times \frac{1}{2}}then\]
Simplify  \(x\times \frac{1}{2}\)  to  \(\frac{x}{2}\).
\[If\times 49={7}^{\frac{x}{2}}then\]
Regroup terms.
\[49If={7}^{\frac{x}{2}}then\]
Regroup terms.
\[49If=ethn\times {7}^{\frac{x}{2}}\]
Divide both sides by \(e\).
\[\frac{49If}{e}=thn\times {7}^{\frac{x}{2}}\]
Divide both sides by \(t\).
\[\frac{\frac{49If}{e}}{t}=hn\times {7}^{\frac{x}{2}}\]
Simplify  \(\frac{\frac{49If}{e}}{t}\)  to  \(\frac{49If}{et}\).
\[\frac{49If}{et}=hn\times {7}^{\frac{x}{2}}\]
Divide both sides by \(h\).
\[\frac{\frac{49If}{et}}{h}=n\times {7}^{\frac{x}{2}}\]
Simplify  \(\frac{\frac{49If}{et}}{h}\)  to  \(\frac{49If}{eth}\).
\[\frac{49If}{eth}=n\times {7}^{\frac{x}{2}}\]
Divide both sides by \(n\).
\[\frac{\frac{49If}{eth}}{n}={7}^{\frac{x}{2}}\]
Simplify  \(\frac{\frac{49If}{eth}}{n}\)  to  \(\frac{49If}{ethn}\).
\[\frac{49If}{ethn}={7}^{\frac{x}{2}}\]
Use Definition of Common Logarithm: \({b}^{a}=x\) if and only if \(log_b(x)=a\).
\[\log_{7}{(\frac{49If}{ethn})}=\frac{x}{2}\]
Multiply both sides by \(2\).
\[\log_{7}{(\frac{49If}{ethn})}\times 2=x\]
Use Power Rule: \(\log_{b}{{x}^{c}}=c\log_{b}{x}\).
\[\log_{7}{{(\frac{49If}{ethn})}^{2}}=x\]
Use Division Distributive Property: \({(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}\).
\[\log_{7}{\frac{{(49If)}^{2}}{{(ethn)}^{2}}}=x\]
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[\log_{7}{(\frac{{49}^{2}{If}^{2}}{{(ethn)}^{2}})}=x\]
Simplify  \({49}^{2}\)  to  \(2401\).
\[\log_{7}{(\frac{2401{If}^{2}}{{(ethn)}^{2}})}=x\]
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[\log_{7}{(\frac{2401{If}^{2}}{{e}^{2}{t}^{2}{h}^{2}{n}^{2}})}=x\]
Switch sides.
\[x=\log_{7}{(\frac{2401{If}^{2}}{{e}^{2}{t}^{2}{h}^{2}{n}^{2}})}\]