working with logs base 2,
since 8=2^3, g(x) = 1/3 log(x-2)
(f+g)(x) = f(x)+g(x)
= 3log(x-2) + 1/3 log(x-2)
= 10/3 log(x-2)
so, if (f+g)(x) = 0,
log(x-2) = 0
x-2 = 1
x = 3
log_2(3-2)^3 = 0
log_8(3-2) = 0
Thanks Steve! I am starting to write these down so that way I will understand them. It is easier for me to go off an example. I appreciate all your help.
math - solve the equation log2(x+4)-log4x=2 the 2 and 4 are lower than the g ...
Logarithms - I'm working on logarithmic equations and I'm stuck on how my book ...
math30 - 1.Use the laws of logarithms to express log2 (6) – log2 (3) + 2log2 (...
Algebra 2 - solve log2(3x-1)-log2(x-1)=log2(x+1) i have absolutely no idea how ...
Math - Logarithmic - Solve: 2^(5x-6) = 7 My work: log^(5x-6) = log7 5x - 6(log2...
Math - Can someone solve this showing the steps involved 4^log2(2^log2 5)
AP Math - Solve: log2(2x^2)[log2(16x)]= 9/2log2x
algebra - Solve for x 1. log5 X=3 2. Log2 16-log2 =x 3. Log9 6561=x
math/algebra - solve the equation log2(3x-2) - log2 (x-5)=4
Precalculus - More log/exponential equations log2 x + log2(x+2) =3