Exponential and Logarithmic Functions

$3^4 = 81$, and a logarithm asks for the exponent.

Adding the indices on top gives $2^8$, then subtracting gives $2^{8-6} = 2^2 = 4$.

Subtracting logs divides their arguments: $\log\dfrac{20}{4} = \log 5$.

Taking logs gives $x\log 2 = \log 20$, so $x = \dfrac{\log 20}{\log 2} = 4.32$. Sensible, since $2^4 = 16$ and $2^5 = 32$.

Change of base: $\log_2 10 = \dfrac{\ln 10}{\ln 2} = 3.32$. It must lie between 3 and 4, since $2^3 = 8$ and $2^4 = 16$.

Combining gives $\log_3\big(x(x+6)\big) = 3$, so $x^2 + 6x = 27$ and $(x+9)(x-3) = 0$. Reject $x = -9$, which would need the log of a negative number.