How do you solve #1.6= log(4x)#?

1 Answer
Gió
Nov 25, 2015

I found: #x=9.9526# for the log in base #10#.

Explanation:

It depends upon the base, #b#, of your log:
#1.6=log_b(4x)#
normally #b# would be #10# because it can be easily evaluated using a pocket calculator; if your base is different, in the next part, simply use your given value instead of #10#.

Use the definition of log:

#log_bx=a->x=b^a#

so that:
#1.6=log_b(4x)# becomes:
#4x=b^(1.6)#
where, if #b=10#, then;
#x=1/4*10^(1.6)#
#x=9.9526#

Related questions
See all questions in Logarithmic Models
Impact of this question
1815 views around the world
You can reuse this answer
Creative Commons License