Math Help Forum: 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.)

do you mean



Is this correct?

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