Please show me step by step on solving this problem? log₅(2x-9)+ log₅(x+2)= 1
Thank you
I'm getting 2x²-5x-28=0 which is a empty set. Trying to make sure I'm correct if I am. Thanks so much for your help
log5 [(2x-9)(x+2)] = 1
(2x-9)(x+2) = 5^1 = 5
2 x^2 -5 x -18 = 5
2 x^2 - 5 x - 23 = 0
x = [ 5 +/- sqrt (25+184 ] /4
= [ 5 +/- 14.5 ] /4
= 19.5/4 or -9.46/4
= 4.86 or -2.36
To solve the equation log₅(2x-9) + log₅(x+2) = 1, you can follow these steps:
Step 1: Combine the two logarithms using the logarithmic identity logₐ(b) + logₐ(c) = logₐ(bc):
log₅[(2x-9)(x+2)] = 1
Step 2: Rewrite the equation using exponential form:
5¹ = (2x-9)(x+2)
Step 3: Simplify the right side of the equation:
5 = 2x² + 4x - 9x - 18
Step 4: Combine like terms:
5 = 2x² - 5x - 18
Step 5: Rearrange the equation to set it equal to zero:
0 = 2x² - 5x - 23
Now, you mentioned that you obtained a quadratic equation of 2x² - 5x - 28 = 0. However, based on the original logarithmic equation, the correct quadratic equation should be 2x² - 5x - 23 = 0.
To solve this quadratic equation, you can use the quadratic formula or factor it if possible.
If you use the quadratic formula, where ax² + bx + c = 0:
x = (-b ± √(b² - 4ac)) / (2a)
For our equation 2x² - 5x - 23 = 0, a = 2, b = -5, and c = -23.
Plugging these values into the quadratic formula, we get:
x = (-(-5) ± √((-5)² - 4(2)(-23))) / (2(2))
x = (5 ± √(25 + 184)) / 4
x = (5 ± √(209)) / 4
Therefore, the solutions are x = (5 + √(209)) / 4 and x = (5 - √(209)) / 4.
To check if these solutions are valid, substitute them back into the original logarithmic equation and see if both sides of the equation are equal.
If you have obtained a different result for the quadratic equation, please double-check your calculations to ensure accuracy.