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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.

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4 Answers

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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.

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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 $a,b > 0, a \neq 1$.

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$\log_a(b) = x \Rightarrow a^x = b$ 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.

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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.

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