If $\log _{ 16 }{ 49 } =a\quad ,\log _{ 7 }{ 2.5 } =b$ then ${ 4 }^{ ab+1 }\quad =\quad ?$
+130081 log16 49 = a log 7 2.5 = b
Using the change of base theorem
ab + 1 =
log 49 log 2.5
_____ * ________ + 1
log 16 log 7
log 49 log 2.5
_____ * _______ + 1
log 7 log 16
log 7^2 log 2.5
______ * ________ + 1
log 7 log 4^2
2log 7 log 2.5
_____ * _______ + 1
log 7 2log 4
2 * log (10/4)
__________ + 1
2log 4
log 10 - log 4
___________ + 1
log 4
log 10 log 4
_____ - ______ + 1
log 4 log 4
1 / log 4 - 1 + 1
1/log 4
So..... using a log property that a^ (1/log a) = 10.....then
4^(ab + 1) = 4^ (1/log 4) = 10
