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

1 Answer
Gió
Dec 7, 2015

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#

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