Answer:
[tex]\frac{9}{32} = \frac{{10}^{2b} }{{10}^{5a}}[/tex]
Step-by-step explanation:
It is given that:
log 2=a and log 3=b
This implies that:
[tex]2 = {10}^{a} [/tex]
and
[tex]3 = {10}^{b} [/tex]
Now we know that:
[tex] \frac{9}{32} = \frac{ {3}^{2} }{ {2}^{5} } [/tex]
We substitute to get:
[tex] \frac{9}{32} = \frac{ ({10}^{b}) ^{2} }{( {10}^{a})^{5} } [/tex]
We simplify to get:
[tex]\frac{9}{32} = \frac{{10}^{2b} }{{10}^{5a}} [/tex]
