Solve (8/2)^2q+1*(3/8)^-5=(27/512)

$$( \frac { 8 } { 2 } ) ^ { 2 q + 1 } \times ( \frac { 3 } { 8 } ) ^ { - 5 } = ( \frac { 27 } { 512 } )$$

Solve for q

$$q=2\log_{2}\left(3\right)-\frac{13}{2}\approx -3.330074999$$

Steps Using Definition of Logarithm

Use the rules of exponents and logarithms to solve the equation.

$$\frac{32768}{243}\times 4^{2q+1}=\frac{27}{512}$$

Divide both sides of the equation by \frac{32768}{243}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$4^{2q+1}=\frac{6561}{16777216}$$

Take the logarithm of both sides of the equation.

$$\log(4^{2q+1})=\log(\frac{6561}{16777216})$$

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

$$\left(2q+1\right)\log(4)=\log(\frac{6561}{16777216})$$

Divide both sides by \log(4).

$$2q+1=\frac{\log(\frac{6561}{16777216})}{\log(4)}$$

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

$$2q+1=\log_{4}\left(\frac{6561}{16777216}\right)$$

Subtract 1 from both sides of the equation.

$$2q=4\log_{2}\left(3\right)-12-1$$

Divide both sides by 2.

$$q=\frac{4\log_{2}\left(3\right)-13}{2}$$