If $\log _{ 16 }{ 49 } =a$ and $\log _{ 7 }{ 2.5 } =b$
then ${ 4 }^{ ab+1 }\quad =\quad ?$
+128408 Note that we can write
log16 7^2 = a (log property log a^b = b log a)
2 log 16 7 = a
log 16 7 = a/2 using the change of base theorem (log a b = log b / log a )
log 7 / log 16 = a/2
2log 7 / log 16 = a
Also
log 2.5 / log 7 = b
So ab = (2log 7)/log 16 * log 2.5 / log 7 = 2log 2.5 / log 16 =
2log 2.5 / log 4^2 = 2.og 2.5 / [ 2log 4 ] = log 2.5 / log 4
So.......
ab + 1 = log 2.5 / log 4 + log4 / log 4 = [ log 2.5 + log 4 ] / log 4 =
(log a + log b = log (a*b) )
log (2.5 * 4) / log 4 =
log 10 / log 4 =
1/log 4
log property a^(1/log a) = 10
4^(1/log 4) = 10
