Solve {c}{text{15.If}(3/4)^6*(16/9)^5=(4/3)^n+2text{,thenthevalueof}xtext{is}}{{llll}{text{(a)}2}&{text{(b)}4}&{text{(c)}-2}&{text{(d)}6}}

$$\left. \begin{array} { c } { \text { 15. If } ( \frac { 3 } { 4 } ) ^ { 6 } \times ( \frac { 16 } { 9 } ) ^ { 5 } = ( \frac { 4 } { 3 } ) ^ { n + 2 } \text { , then the value of } x \text { is } } \\ { \left. \begin{array} { l l l l } { \text { (a) } 2 } & { \text { (b) } 4 } & { \text { (c) } - 2 } & { \text { (d) } 6 } \end{array} \right. } \end{array} \right.$$

Solve for n

$$n=2$$

Steps Using Definition of Logarithm

Calculate \frac{3}{4} to the power of 6 and get \frac{729}{4096}.

$$\frac{729}{4096}\times \left(\frac{16}{9}\right)^{5}=\left(\frac{4}{3}\right)^{n+2}$$

Calculate \frac{16}{9} to the power of 5 and get \frac{1048576}{59049}.

$$\frac{729}{4096}\times \frac{1048576}{59049}=\left(\frac{4}{3}\right)^{n+2}$$

Multiply \frac{729}{4096} and \frac{1048576}{59049} to get \frac{256}{81}.

$$\frac{256}{81}=\left(\frac{4}{3}\right)^{n+2}$$

Swap sides so that all variable terms are on the left hand side.

$$\left(\frac{4}{3}\right)^{n+2}=\frac{256}{81}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{4}{3}\right)^{n+2})=\log(\frac{256}{81})$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$\left(n+2\right)\log(\frac{4}{3})=\log(\frac{256}{81})$$

Divide both sides by \log(\frac{4}{3}).

$$n+2=\frac{\log(\frac{256}{81})}{\log(\frac{4}{3})}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$n+2=\log_{\frac{4}{3}}\left(\frac{256}{81}\right)$$

Subtract 2 from both sides of the equation.

$$n=4-2$$