SOLUTION: Solve for x : log4 (x^2) = log2 (2x-1) A large part of my difficulty with this one is that I think most calculators have Log10 set as the default log, and I don't know how to c

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve for x : log4 (x^2) = log2 (2x-1) A large part of my difficulty with this one is that I think most calculators have Log10 set as the default log, and I don't know how to c      Log On


   

Question 987574: Solve for x : log4 (x^2) = log2 (2x-1)
A large part of my difficulty with this one is that I think most calculators have Log10 set as the default log, and I don't know how to change that to do this equation.
Answer by ikleyn(52797) About Me  (Show Source):
You can put this solution on YOUR website!
.
Your equation is

log%284%2C%28x%5E2%29%29 = log%282%2C+%282x-1%29%29.

It is not about using a calculator.  It is about solving the equation.

The domain is  2x - 1 > 0,  or  x > 1%2F2.

First,  you have  log%284%2C%28x%5E2%29%29 = 1%2F2log%282%2C%28x%5E2%29%29,  since  4 = 2%5E2.

Substitute it into the equation.  You will have

1%2F2log%282%2C%28x%5E2%29%29 = log%282%2C+%282x-1%29%29,

log%282%2C%28x%5E2%29%29 = 2log%282%2C+%282x-1%29%29,

log%282%2C%28x%5E2%29%29 = 2log%282%2C+%282x-1%29%29%5E2.

Hence,

x%5E2 = %282x-1%29%5E2,

x%5E2 = 4x%5E2+-+4x+%2B+1,

3x%5E2+-+4x+%2B+1 = 0.

x%5B1%2C2%5D = %284+%2B-+sqrt%2816+-12%29%29%2F6 = %284+%2B-+2%29%2F6.

x%5B1%5D = %284%2B2%29%2F6 = 1.     It is in the domain.

x%5B2%5D = %284-2%29%2F6 = 1%2F3.     It is outside the domain.

Answer.  There is the unique solution  x = 1.