# Math Help - Another Logarithmic question

1. ## Another Logarithmic question

Solve for x.

a) 2^x+2-2^x =96

2^x (2^2 +2) =96 nvm i did addition instead of subtraction i see my mistake... on this question.

2^x(6)=96/6

2^x=16

2^4=16

x=4 (the answer at the back of the book says its 5)

c) 4^x+1 +4^x=160

4^x (4^2+1)=160

4^x(17)=160/17

4^x=9.41

(the answer is suppose to be 2.5 but i dont see where they are coming from.)

2. do you mean

$2^x+2-2^x =96$

Is this correct?

3. Originally Posted by Skoz
Solve for x.

a) 2^x+2-2^x =96

2^x (2^2 +2) =96 nvm i did addition instead of subtraction i see my mistake... on this question.

2^x(6)=96/6

2^x=16

2^4=16

x=4 (the answer at the back of the book says its 5)

c) 4^x+1 +4^x=160

4^x (4^2+1)=160

4^x(17)=160/17

4^x=9.41

(the answer is suppose to be 2.5 but i dont see where they are coming from.)
please use parentheses to make your math expressions/equations clear, like so ...

2^(x+2) - 2^x = 96

4^(x+1) + 4^x = 160

$4^{x+1} + 4^x = 160$

$4^x(4 + 1) = 160$

$4^x = 32$

$2^{2x} = 2^5$

$2x = 5$

$x = 2.5$