Logarithm help please

Question

Solve Log_bx=2-a+log_b(a²b^a/b²)

Answer 1

First, try to simplify the equation using log(xy)= log x + log y.

Log_bx=2-a+log_b(a²b^a/b²), which is equivalent to Log_bx=2-a+log_b(a²)+log_b(b^a/b²), and so we can get



Log_bx = 2-a+log_b(a²)+log_b(b^a-2),

Log_bx = 2-a +log_b(a²)+(a-2),

Log_bx = log_b(a²)

x = b^(log_b(a²))

x= a².

Here sometimes I had to use the notation ^, which is the same thing as power.

For example: a^2 = a².

Answer 2

First the log on the right

log_b (a²b^a / b²)

log_b a² + log_b b^a - lob_b b²

2 log_b a + a - 2

Add the "2 - a"

2 log_b a

The equation becomes

log_b x = 2 log_b a

log_b x = log_b a²

x = a²