Question

How do you solve `log_(121)44 = x`?

Answer 1

I found: `x=0.789`

Explanation:

We can write (using the definition of log):
`44=121^x`
then
`11*4=11^(2x)`
taking `11` to the right:
`4=11^(2x)/11`
using the property of the quotient of exponents with the same base:
`4=11^(2x-1)`
now we can take the natural log of both sides:
`ln(4)=ln(11^(2x-1))`
we can now use the property of the logs:
`logx^a=alogx`
to get:
`ln4=(2x-1)*ln11`
so that:
`2x-1=ln(4)/ln(11)`
rearranging:
`x=1/2[ln(4)/ln(11)+1]=0.789`