Friday

January 30, 2015

January 30, 2015

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

log2(log4x)=1 and

solve for x and y:

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

- Math -
**helper**, Sunday, January 23, 2011 at 11:41pmlog2(log4x)=1

Rewrite as

(log((log(x))/(log(4))))/(log(2)) = 1

Divide both sides by 1/(log(2)):

log((log(x))/(log(4))) = log(2)

Cancel logarithms by taking exp of both sides:

(log(x))/(log(4)) = 2

Divide both sides by 1/(log(4)):

log(x) = 2 log(4)

Cancel logarithms by taking exp of both sides:

x = 16

You need to rewrite the 2nd with parentheses.

As written it is not clear.

(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 ...

Math - Can someone solve this showing the steps involved 4^log2(2^log2 5)