Question
I need help, can someone guide me through this problem?
log3(x + 2) + log3(4x – 1) = 5
log3(x + 2) + log3(4x – 1) = 5
Answers
Damon
sum of logs is log of product
log3 (4 x^2 + 7 x -2) = 5
base ^logbase(z) = z
so
3^[log3 (4 x^2 + 7 x -2)] = 3^5
or
(4 x^2 + 7 x -2) = 243
4 x^2 + 7 x - 245 = 0
(x-7)( 4x+35) = 0
x = 7 or x = -35/4
log3 (4 x^2 + 7 x -2) = 5
base ^logbase(z) = z
so
3^[log3 (4 x^2 + 7 x -2)] = 3^5
or
(4 x^2 + 7 x -2) = 243
4 x^2 + 7 x - 245 = 0
(x-7)( 4x+35) = 0
x = 7 or x = -35/4
Me
thank u so much!!! honestly u don't know how much u helped me :)
Damon
You are welcome, good luck :)