Question

$$3 ^ { 4 } \times 9 ^ { 2 } = 3 ^ { n } n = ?$$

Answer

$$h=6561/(e*t*n^2*3^n)$$

Solution


Simplify  \({3}^{4}\)  to  \(81\).
\[81\times {9}^{2}={3}^{n}thenn\]
Simplify  \({9}^{2}\)  to  \(81\).
\[81\times 81={3}^{n}thenn\]
Simplify  \(81\times 81\)  to  \(6561\).
\[6561={3}^{n}thenn\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[6561={3}^{n}the{n}^{2}\]
Regroup terms.
\[6561=eth{n}^{2}\times {3}^{n}\]
Divide both sides by \(e\).
\[\frac{6561}{e}=th{n}^{2}\times {3}^{n}\]
Divide both sides by \(t\).
\[\frac{\frac{6561}{e}}{t}=h{n}^{2}\times {3}^{n}\]
Simplify  \(\frac{\frac{6561}{e}}{t}\)  to  \(\frac{6561}{et}\).
\[\frac{6561}{et}=h{n}^{2}\times {3}^{n}\]
Divide both sides by \({n}^{2}\).
\[\frac{\frac{6561}{et}}{{n}^{2}}=h\times {3}^{n}\]
Simplify  \(\frac{\frac{6561}{et}}{{n}^{2}}\)  to  \(\frac{6561}{et{n}^{2}}\).
\[\frac{6561}{et{n}^{2}}=h\times {3}^{n}\]
Divide both sides by \({3}^{n}\).
\[\frac{\frac{6561}{et{n}^{2}}}{{3}^{n}}=h\]
Simplify  \(\frac{\frac{6561}{et{n}^{2}}}{{3}^{n}}\)  to  \(\frac{6561}{et{n}^{2}\times {3}^{n}}\).
\[\frac{6561}{et{n}^{2}\times {3}^{n}}=h\]
Switch sides.
\[h=\frac{6561}{et{n}^{2}\times {3}^{n}}\]