# Maths

posted by .

I got this question, and found what I believe to be the solution, but want it confirmed.

I started with:
log[base5](2x+1) + log[base5](x-1) = 1

And used the product law:
log[base5]((2x+1)(x-1)) = 1
log[base5](2x^2-x-1) = 1

Then I changed to exponential form:
5^1 = 2x^2-x-1
5 = 2x^2-x-1
5-5 = 2x^2-x-1-5
0 = 2x^2-x-6

Then, to find x I used the quadratic formula, ending up with:

x = (1 [+ or -] 5) / 4

The second value of x, -1, was inadmissable, so my final value for x was 3/2.

Is that correct?

• Maths -

Your equation up to
2x^2-x-6 = 0 is correct, but I factored it to get

(2x+3)(x-2)
so x = 2 or x = -3/2 which is inadmissable

check:
if x=2
LS = log5 5 + log51
= 1+0 = 1 = RS

so x = 2

• Maths -

Thank you :) One question, however. Where did you get LS = log[base5]5 + log[base5]1?

At the beginning, LS = log[base5](2x+2) + log[base5](x-1)

Sorry, it just confused me, it would help if you'd explain that.

• please come back... i still have questions -

Please? ^

### Respond to this Question

 First Name School Subject Your Answer

### Similar Questions

More Related Questions

Post a New Question