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
so x = 2 or x = -3/2 which is inadmissable
LS = log5 5 + log51
= 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
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