Question

solve this equation ln(2x+5)+ln(x-4)-2lnx=0

Answer 1

log(A) + log(B) - log(C) = log(AB/C).
Write 0 as log(1)
Equate the two and solve for x.

Note if log(P) = log(Q) then P = Q.

Answer 2

im so confused

Answer 3

what does x equal rounded to 2 decimal places

Answer 4

\[
\ln(2x+5) + \ln(x-4) - 2\ln(x) = 0 \\
\ln\{(2x+5) * (x-4)\} - \ln(x^2) = 0 \\
\ln\{(2x+5) * (x-4)/x^2\} = 0 \\
\ln\{(2x+5) * (x-4)/x^2\} = \ln(1) \\
\{(2x+5) * (x-4)/x^2\} = 1 \\
(2x+5)(x-4) = x^2
\]Solve for x

Answer 5

i dont think this is right

Answer 6

i need to end up with 1 answer

Answer 7

look right to me

Answer 8

When you solve the quadratic you will get 2 values for x.
But you have to test each value in the original problem to make sure we don't end up taking log of a number less than or equal to 0. If so discard that x value as an extraneous solution.

Answer 9

i got root 6 and i root 6 over 2

Answer 10

im supposed to have a decimal

Answer 11

(2x+5)(x-4) = x^2
2x^2 - 8x + 5x - 20 = x^2
x^2 - 3x - 20 = 0
x = [ 3 +/- sqrt{9 + 80} ] / 2
x = 1/2 * (3 + sqrt(89)} or x = 1/2 * (3 - sqrt(89)}
The second one is negative and we can discard it because we can't take log of a negative number.
So x = 1/2 * (3 + sqrt(89)} = 6.22 (to two decimal places).