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Let a,b be positive real numbers that satisfy

                                      2 + log2(a) = 3 + log3(b) + log6(a+b)

 

Find (a+b)/ab

 Jul 27, 2020
 #1
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(a + b)/(ab) = 8.

 Jul 27, 2020
 #2
avatar+110235 
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Are 2,3 and 6 bases?

 

I think 2 3 and 6 are not bases but factors. But then i need to know what base the question is in.

I think you have left out key information.

 Jul 27, 2020
edited by Melody  Jul 27, 2020
 #3
avatar+110235 
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Let me assume base 10

 

\(2+log(2a)=3+log(3b)+log[6(a+b)]\\ log2+log a=1+log3+logb + log6 + log(a+b) \\ log2 - log3 - log6 -1= logb + log(a+b) - log a \\ log2 - log3 - log6 -log10= log(a+b) - log a + logb \\ log\frac{2}{3*6*10} = log\left[\frac{(a+b)}{a} *b\right] \\ \frac{2}{3*6*10} = \frac{(a+b)}{a} *b\\~\\ \frac{(a+b)}{a} *b=\frac{1}{90} \)

 Jul 27, 2020

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