SOLUTION: I need to solve for x in the equation: log base 7 of 4x= log of sqrt of x I began with changing the sqrt to 1/2 and moving it in front of the right side of the equation. I then

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I need to solve for x in the equation: log base 7 of 4x= log of sqrt of x I began with changing the sqrt to 1/2 and moving it in front of the right side of the equation. I then      Log On


   

Question 810984: I need to solve for x in the equation:
log base 7 of 4x= log of sqrt of x
I began with changing the sqrt to 1/2 and moving it in front of the right side of the equation. I then pulled the right side of the equation to the left as an exponent, which gave me:
7^(1/2logx)=4x
I then tried to make another exponent in:
7^(1/2(10^4x))=x
am I at all on the right track?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The way you started had a sign that said "No outlet" and you ended up trapped in a cul de sac.
It was not an illegal "Wrong way",
and up to 7%5E%28%281%2F2%29log%28x%29%29=4x you had not done anything wrong,
but it was not the way that would get you to the answer.
You were expected to deal with the different logarithm bases by doing a change of base.
It's like converting everything to the same units as soon as possible when they give you mixes units in a problem.

For log%287%2C4x%29=log%28%2810%29%2Cx%29 , I would change bases to have everything on base 10.
It is a popular base.
You can get base 10 logarithms from calculators and tables,
and, by convention, you can skip writing the base.

log%287%2C4x%29=log%28x%29-->log%284x%29%2Flog%287%29=log%28x%29
I do not need to write bases any more.
log%284x%29%2Flog%287%29=log%28x%29-->log%284x%29=log%28x%29%2Alog%287%29-->log%284%29%2Blog%28x%29=log%28x%29%2Alog%287%29-->log%284%29=log%28x%29%2Alog%287%29-log%28x%29-->log%284%29=log%28x%29%2A%28log%287%29-1%29-->log%28x%29=log%284%29%2F%28log%287%29-1%29
You can re-write that expression in other ways:

I cannot think of an elegant way to express the answer as an exact answer.
x=10%5E%28log%280.7%2C4%29%29 does not look elegant,
but x=10%5E%22%28log%284%29%2F%28log%287%29-1%29%29%22 does not look that much better.
However, based on log%28x%29=log%284%29%2F%28log%287%29-1%29 or log%28x%29=log%284%29%2Flog%280.7%29
we can calculate an approximate answer as
x=0.0001298 .