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
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.
mathematics - Need help with these: Write the expression as a logarithm of a ...
Math: Algebra2 Logarithms - Can someone help me find the value of B in this ...
Math - log(base5)x+log(base25)x+log(base125)x=33 solve for x
Algebra 2 - What is X? log(base 5)x^10-log(base5)x^6=21
math - find the inverse of f(x)=-5+log(base5)(y-5)
pre calculus - Slove: 2logbase5(x+2) = 2 + log base5(x-2)
Algebra 2 - how do i solve these two problems? 3log(base5)^(x^2+9)-6 = 0 log(...
Algebra 2 - how do not understand how to do this Log X + Lob (X-3) = 1 I know I ...
Math - Can the product law of logarithms be used to evaluate log(-2) + log(-3)? ...
Math Help Please - Which of the following expressions is equal to log (x sqrt-y...