How do you solve #log_5(3x^2 - 1) = log_5 2x#?

Question

Anna Allen · Accepted Answer

x = 1.

The logarithm function only has one value. So, #3x^2-1=2x#. Solving this quadratic equation #3x^2-2x-1=0, x = 1. -1/2# Negative root is inadmissible for #log_5(2x). Thus, x = 1.

Elias Ackerman · Answer

undefined To solve the equation log_5(3x^2 - 1) = log_5 2x, we can use the property of logarithms that states if log_b(x) = log_b(y), then x = y. Therefore, we can set the expressions inside the logarithms equal to each other:

3x^2 - 1 = 2x

Next, we bring all terms to one side of the equation:

3x^2 - 2x - 1 = 0

Now, we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:

x = [-b ± √(b^2 - 4ac)] / (2a)

Where a = 3, b = -2, and c = -1:

x = [2 ± √((-2)^2 - 4(3)(-1))] / (2 * 3)

x = [2 ± √(4 + 12)] / 6

x = [2 ± √16] / 6

x = [2 ± 4] / 6

This gives us two possible solutions:

x = (2 + 4) / 6 = 6 / 6 = 1

x = (2 - 4) / 6 = -2 / 6 = -1/3

Therefore, the solutions to the equation log_5(3x^2 - 1) = log_5 2x are x = 1 and x = -1/3.