Posted by **anonymous** on Tuesday, June 8, 2010 at 6:22pm.

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 -
**Reiny**, Tuesday, June 8, 2010 at 8:10pm
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 = log_{5} 5 + log_{5}1

= 1+0 = 1 = RS

so x = 2

- Maths -
**anonymous**, Tuesday, June 8, 2010 at 8:29pm
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 -
**anonymous**, Tuesday, June 8, 2010 at 8:43pm
Please? ^

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