Posted by anonymous on Tuesday, January 8, 2013 at 6:30am.
we should have the same base in the logs
let log4x = y
then 4^y = x
2^(2y) = x
log2 x = 2y
y = (1/2)log2 x
= log2 x^(1/2)
= log2√x
then log4x = log2 √x
log2[ (x+4)/√x] = 2
(x+4)/√x = 2^2 = 4
x+4 = 4√x
square both sides
x^2 + 8x + 16 = 16x
x^2 - 8x + 16 = 0
(x-4)^2 = 0
x-4 = 0
x = 4
Related Questions
Math - I am having trouble figuring out how to solve these logarithms. Could ...
Math - I am having trouble figuring out how to solve these logarithms. Could ...
Math-Advanced Functions - I am having trouble figuring out how to solve this ...
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 - 2. Simplify. Please be sure to show all of your work. -3(-9) |-5...
Math - log4x^4 + log4^2 = 3 Solve the logarithmic equation.
Math Log - How do you find log20, if log2=.301?? Please show work
Algebra - Solve the logarithmic equation. log2(4x)-log2(x+3)=1
precalculus - find the exact value of the following logarithm. Use the ...
For Further Reading