Solve the equation log5(x) = 16logx(5)
I was just stuck on these two questions, how do I solve for the variable?
The trick here is to use the change of base law \(\log_b(a)=\frac{\log_c a}{\log_c b}\) to obtain everything in the same base. Note that I've just used \(\log\) without specifying any base as the specific base is
irrelevant.
\begin{align}
\log_5(x) &= 16\log_x(5) \\
\dfrac{\log x}{\log 5} &= 16\left(\dfrac{\log5}{\log x} \right)\\
(\log x)^2 &= 16(\log5)^2\\
\log x&=\pm 4\log 5\\
\log x&=\log 5^{4},\log 5^{-4}\\
x&= 5^{4},5^{-4}
\end{align}
Given that q^p=25, find log5(q) in terms of p
Solve for q in terms of p, then substitute that into the given expression.
\begin{align}
q^p=25 &\implies q=25^{1/p}\\
&\implies \log_5(q)=\log_5(25^{1/p})=\frac{1}{p}\log_5(25)=\frac{2}{p}
\end{align}