Posted by **Chloe** on Sunday, January 23, 2011 at 11:20pm.

I am having trouble figuring out how to solve these logarithms. Could someone please help!

log2(log4x)=1 and

solve for x and y:

(1/2)^x+y= 16 logx-y8=-3

## Answer this Question

## Related Questions

- 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 ...
- math - I tried everthing but,I couldnt solve it. Please help me.the answer at ...
- math - solve the equation log2(x+4)-log4x=2 the 2 and 4 are lower than the g ...
- math - solve the equation. log2(x+4)-log4x=2 please show work
- logarithms - how do i solve: log4x^3 + log2x^1/2 = 8 please help me
- MATH - Solve: logx^3 - log2 = log(2x^2)
- College Math II - Solve by any method. a^4 – 5a^2 + 4 = 0 I am having problems ...
- Algebra 2 - solve log2(3x-1)-log2(x-1)=log2(x+1) i have absolutely no idea how ...
- maths - 1. Log10²x+log10x²=log10² 2-1 2. Log4(log2x)+log2(log4x)=2 3. X^logx+5/3...

More Related Questions