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If a, b and c are positive numbers, than equality $$\log_{a} c + \log_{b} c = \log_{a+b} c$$ is true if and only if $$1 + \log_{b} a = \log_{a+b} a$$ Prove it!

I have looked at the solution but it is not clear for me.

We will prove that if $$\log_{a} c + \log_{b} c = \log_{a+b} c$$ than it is $$1 + \log_{b} a = \log_{a+b} a$$

$$\log_{a} c + \log_{b} c = \frac{\log_{a+b} c}{\log_{a+b} a} +\frac{\log_{a+b} c}{\log_{a+b} b} =\log_{a+b} c $$

So this is only thing that is not clear for me how is this equal. $$\frac{\log_{a+b} c}{\log_{a+b} a} +\frac{\log_{a+b} c}{\log_{a+b} b} =\log_{a+b} c $$

Ivan Ljilja
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  • That looks like the original equation using a change of base formula for logarithms. – Michael Burr May 01 '20 at 17:00
  • Note that the numerator cancels from both sides, so suffices to show that $$ \frac{1}{\log_{a+b} a} + \frac{1}{\log_{a+b} b} = 1 $$ but this seems false. E.g. $a=b=e/2$ you get $$ 1 = \frac{1}{\ln(e/2)} + \frac{1}{\ln(e/2)} = \frac{2}{\ln(e/2)} = \frac{2}{1 - \ln 2} $$ which is clearly not true since the RHS is approximately $6.5$... – gt6989b May 01 '20 at 17:10
  • You are trying to prove that $\log_{a} c + \log_{b} c = \log_{a+b} c$ when you want to prove that if $\log_{a} c + \log_{b} c = \log_{a+b} c$, then $1 + \log_{b} a = \log_{a+b} a$. – mathlove May 01 '20 at 17:54

2 Answers2

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Note that

$$log_b (a)= \frac{ln (a)}{ln(b)}$$ where $ln(a)$ is the natural log of $a$.

Any arbitrary base can be assumed instead of $e$.

So, if

$$log_a (c)+ log_b (c)= log_{a+b} (c) $$

$$\implies\frac{ln(c)}{ln(a)}+\frac{ln(c)}{ln(b)}=\frac{ln(c)}{ln(a+b)}$$ $$\implies\frac{1}{ln(a)}+\frac{1}{ln(b)}=\frac{1}{ln(a+b)}\ ,\ \ assuming \ \ c\neq1$$ $$\implies1+\frac{ln(a)}{ln(b)}=\frac{ln(a)}{ln(a+b)}$$ $$\implies1+log_b(a)=\frac{ln(a)}{ln(a+b)}$$ $$\implies1+log_b(a)=log_{a+b}(a)$$

The steps are perfectly reversible.

Shooter
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WE have:

$$1+ log_b (a)= log_{a+b} (a) $$

Let $a=c$ then:

$$ log _a(c)= log _c (c)=1$$

therefore:

$$log _a (c) + log _b (c)= log _{a+b} (c)$$

sirous
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