Power = integer problems

Question

Solving for $x$ for an exact answer: $5^{x+1} = 41$

My first attempt was as follows:
$5^{x+1} = 41,$
thus $\log_5 (41) = x+1,$
thus $x = \log_5 (41) - 1$

Question: Is this answer considered not correct because of the base $5$ in the logarithm, which should be $10$?

Second attempt: $5^{x+1} = 41,$
thus $\log(5^{x+1}) = \log41,$
thus $(x+1)(\log5) = \log41,$
thus $x\log5+\log5 = \log41,$
thus $x\log5 = \log41-\log5,$
thus $x = {\log41-\log5\over \log5}$

Third attempt: $5^{x+1} = 41,$
thus $\log(5^{x+1}) = \log41,$
thus $(x+1)(\log5) = \log41,$
thus $(x+1) = {\log41\over \log5},$
thus $x = {\log41\over \log5} - 1$

Question: Why do we get two different answers in my second and third attempts? I understand the third way is optimal, but what are wrong about the third (and first) ways?

Thanks!

Answer 1

All answers are true.

$$\frac{a-x}{x}=\frac{a}{x}-1\\\log_ab=\frac{\log b}{\log a}$$

Answer 2

All your answers are correct and equal. My calculator returns $1.307372057$ for each of the answers you got. It is just a matter of simplification that is causing the differences.

For example, in attempt #2, you have $\frac{\log {41}-\log5}{\log 5}$. It is a rule in basic algebra that you could write it as: $\frac{\log{41}}{\log 5} - \frac{\log 5}{\log5}$. You know that for a nonzero number $a$, it is true that $\frac aa=1$. So your answer becomes $\frac{\log{41}}{\log 5}-1$, which is your third attempt.

But also, some mathematicians prefer the natural logarithm ($\ln$), but I won't talk about this.