Question

PLEASE HELP!!
IM STUCK ....

Answer 1

$$
\log_a(x-2) + \log_a(2x-1) = 2 \log_a(x) \\
\log_a \Big[ (x-2)(2x-1) \Big] = \log_a \left(x^2 \right)
$$
At this point you it is important to know the exact domain of your log function.
If we assume it could only return real numbers, then it can only work for positive numbers.
In that case we have to check for both:
$$
(x-2)(2x-1) > 0 \\
x^2 > 0
$$
After we found the values of x where the operation is defined we can take both sides to the power of 'a' and get rid of the log:
$$
(x-2)(2x-1) = x^2 \\
(2x^2 -5x + 2) - x^2 = 0 \\
x^2 -5x + 2 = 0
$$Now we have simple quadratic equation. after we solve it we have to compare the solutions of x to the possible values we found for it above and see what we're left with

Answer 2

where did u get x^2 in the last step??

Answer 3

@pitamar

Answer 4

$$
\log_a(x-2) + \log_a(2x-1) = 2 \log_a(x) \\
\log_a \Big[ (x-2)(2x-1) \Big] = \log_a \left(x^2 \right) \\
a^\left( \log_a \Big[(x-2)(2x-1)\Big] \right) = a ^ \left( \log_a(x^2) \right) \\
(x-2)(2x-1) = x^2 \\
(x-2)(2x-1) - x^2 = 0 \\
(2x^2 - 5x + 2) - x^2 = 0 \\
x^2 -5x + 2 = 0
$$