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?
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
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? ^
Yes, your solution is correct. Let's go through the steps to confirm it.
Starting with the equation:
log[base5](2x+1) + log[base5](x-1) = 1
You correctly used the product law of logarithms to combine the two logarithms:
log[base5]((2x+1)(x-1)) = 1
Then, you changed the equation to its exponential form:
5^1 = (2x+1)(x-1)
5 = 2x^2 - x - 1
Next, you simplified the equation by subtracting 5 from both sides:
0 = 2x^2 - x - 6
To solve for x, you used the quadratic formula:
x = (-b +/- sqrt(b^2 - 4ac)) / 2a
In this case, a = 2, b = -1, and c = -6. Substituting these values into the quadratic formula, you get:
x = (1 +/- sqrt((-1)^2 - 4(2)(-6))) / (2(2))
x = (1 +/- sqrt(1 + 48)) / 4
x = (1 +/- sqrt(49)) / 4
x = (1 +/- 7) / 4
Here, you correctly noticed that the second value of x, -1, is inadmissible because it makes the logarithms undefined. So the only valid solution is:
x = (1 + 7) / 4
x = 8 / 4
x = 2
Therefore, your final value of x as 3/2 is incorrect. The correct solution is x = 2.