Solve for x in the equation :)

Question

log_0.1(3x+5) + logx²/5= -1

Answer 1

First, you need to change the logs to the same base. Let's change the log_0.1(3x+5) to base 10 (like the other log). Use the change of base formula:

log_0.1(3x+5) = log(3x+5)/log(0.1) = log(3x+5)/log(10^-1) = log(3x+5)/(-1) = -log(3x+5)

Now substitute the result into the original equation:

-log(3x+5) + log(x²/5) = -1

log(3x+5) - log(x²/5) = 1 [Multiplied both sides by -1]

log((3x+3)/(x²/5)) = 1 [Log Property: log(a) - log(b) = log(a/b)]

log(5(3x+5)/x²) = 1

5(3x+5)/x² = 10¹

3x + 5 = 2x²

0 = 2x² - 3x - 5

Factor the quadratic using the ac method and solve for x to get the answers.

Answer 2

log_0.1(3x+5) + log(x²/5) = -1

log(3x+5) / log(0.1) + log(x²/5) = -1

log(x²/5) - log(3x+5) = -1

log[x² / (15x+25)] = -1

x² / (15x+25) = 1/10

10x² = 15x + 25

10x² - 15x - 25 = 0

2x² - 3x - 5 = 0

(2x - 5)(x + 1) = 0

x = 5/2 or -1