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I have just started learning logarithms and I am struggling to solve the following

$$ \begin{equation} 2^a3^b = 6 \\ 3^a4^b = 6 \end{equation} $$

How can we solve for a and b in this case?

EDIT 1: Cleared up the two equations to increase understanding. Thanks for the heads-up!

abracadebby
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    This is not clear. Are you trying to write two separate equations with different $a,b$? Of course one of these equations like $2^a3^b=6$ has infinitely many solutions. – lulu Feb 21 '20 at 13:00
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    If you were to take the logarithm of everything, you'll have $\log(2^a3^b) = \log(3^a4^b)=\log(6)$. Now, looking at the first more closely, remember that multiplication inside of a logarithm becomes addition outside, and exponents inside become factors outside, so the first expression can simplify as $\log(2^a3^b)=a\log(2)+b\log(3)$. Similarly the other can simplify leaving you with $a\log(2)+b\log(3)=a\log(3)+b\log(4)=\log(6)$. Can you continue from here? Don't be afraid of $\log(2),\log(3),\dots$ they are just numbers like any other. – JMoravitz Feb 21 '20 at 13:07
  • yep they're two different equations thanks for clarifying! – abracadebby Feb 21 '20 at 13:07
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    If it helps to simplify things in the end, it might be worth noting that $\log(4)=2\log(2)$ and that $\log(6)=\log(2)+\log(3)$ – JMoravitz Feb 21 '20 at 13:08
  • @JMoravitz I see! That makes a lot of sense! thank you so much for that tip! :D – abracadebby Feb 21 '20 at 13:08
  • @JMoravitz Apparently, you were faster :) – Peter Feb 21 '20 at 13:09
  • The solution is actually unique since the linear equation system we get has a unique solution. – Peter Feb 21 '20 at 13:11

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Hint : Taking the natural logarithm on both sides , you get the linear equation system $$\ln(2)\cdot a+\ln(3)\cdot b=\ln(6)$$ $$\ln(3)\cdot a+\ln(4)\cdot b=\ln(6)$$ which should be easy to be solved.

Peter
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