Value of $x$ when $5 + \log x = \log \left(x^6\right)$

Question

Find the value of $x$ when $$5 + \log x = \log \left(x^6\right)$$

I've tried many times to solve this, however I can't seem to find a correct (consistent) answer. My solutions range from $$x = e, x = \sqrt 5, x = \sqrt 6$$

Answer 1

\begin{align} 5+\log x&=\log x^6\\ \Longrightarrow 5+\log x&=6\cdot\log x\\ \Longrightarrow 5&=6\log x-\log x\\ &=5\log x\\ \Longrightarrow 1&=\log x \end{align} Your answer from here depends on what base you have for your log function, and if we assume that it is to the base $e$, then $x=e$.

Answer 2

$$5\ln e+\ln x=\ln x^6\rightarrow \ln e^5+\ln x=\ln x^6\rightarrow\ln (xe^5)=\ln x^6\rightarrow xe^5=x^6 \rightarrow e^5=x^5\rightarrow ~~x=e$$

Answer 3

$$5+\log x=\log x^6\implies 5+\log x=6\log x\\\implies 5=5\log x\implies 1=\log x\implies x=e$$, assuming log is the natural log.