+0

# LOG PROBLEM

+1
76
3
+97

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
0

(a + b)/(ab) = 8.

Jul 27, 2020
#2
+111082
+1

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
+111082
+1

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