Question

Log5 (2x+1)+log5 (16x-7)=3

Answer 1

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Answer 2

To solve the equation log₅(2x+1) + log₅(16x-7) = 3, we can use the properties of logarithms to simplify it.

The sum of logarithms can be expressed as the logarithm of the product, so we have:

log₅[(2x+1)(16x-7)] = 3

Now, let's apply the logarithm's definition:

₅³ = (2x+1)(16x-7)

125 = (2x+1)(16x-7)

Expand the equation:

125 = 32x² - 14x + 16x - 7

Combine like terms:

125 = 32x² + 2x - 7

Rearrange the equation to set it equal to zero:

32x² + 2x - 7 - 125 = 0

32x² + 2x - 132 = 0

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 32, b = 2, and c = -132. Substituting these values into the quadratic formula, we get:

x = (-2 ± √(2² - 4(32)(-132))) / (2(32))

Simplifying further:

x = (-2 ± √(4 + 16896)) / 64

x = (-2 ± √16900) / 64

x = (-2 ± 130) / 64

We have two possible solutions:

1. x = (-2 + 130) / 64 = 128 / 64 = 2

2. x = (-2 - 130) / 64 = -132 / 64 = -33/16

Therefore, the solutions to the equation are x = 2 and x = -33/16.