Prove that $\log_a(b)=\log(b)/\log(a)$ Prove that $$\log_a(b)=\log(b)/\log(a)$$
I don't know how to solve it, but I need to prove it so solve a problem.
 A: Hint: yours holds if and only if
$$ \log(a) \log_a(b) = \log(b)$$
which in turn is true if and only if
$$ e^{\log(a) \log_a(b)} = e^{\log(b)} \ldotp$$
I hope you can take it from here, if not post what you got and we'll see further.
A: We have
$$
a^{\log_a(b)} = b
$$
so we need to show that this holds for the right-hand side as well:
$$
a^{\log(b)/\log(a)}\\
= (10^{\log(a)})^{\log(b)/\log(a)}\\
= 10^{\log(a) \cdot \log(b)/\log(a)}\\
= 10^{\log(b)}\\
= b
$$
Since $a$ raised to one of them is the same as $a$ raised to the other one, they must be equal. This is all the time assuming that $a,b > 0, a \neq 1$.
A: $\log_a(b) = x \Rightarrow a^x = b$
Now taking logarithms of both sides gives
$$\log(a^x) = \log(b) \Rightarrow x\log(a) = \log(b) \Rightarrow x = \frac{\log(b)}{\log(a)}$$
But $x=\log_a(b)$ so we are done.
A: Consider the following :
Let $loga(b)= m$ ...(Assumption)
Therefore, $b=a^m$ ...(by definition of log)
Now look at the Right Hand Side of the required proof,
Let $log(b)/log(a) = n$
Substitute $b=a^m$ in it..
Therfore, $log(b)=log(a^m)=mlog(a)$ ...(Property of log)
Hence $logb/loga=mloga/loga=m$
Hence $m=n$
Thus required result is proved.
