2^x + 1/2^x = 5
let y = 2^x
y + 1/y = 5
y^2 + 1 = 5 y
y^2 -5y +1 = 0
y = [5 +/- sqrt(21) ] /2
y = [2.5 +/- 2.3] approximately
y = 4.8 or y = .2
if 2^x = 4.8
x log 2 = log 4.8 solve for x
if y = .2, same deal,perhaps same answer
Solve the Equation:
1. 2^x + 2^-x = 5
2. Log2x + log2(4 – x) = 0
thanks. i'm not sure how to approach this at all.
2 answers
Now on this one I am not sure what you mean.
log base 2 (x) or log (2x)
I will assume base 2
then
log x + log (4-x) = 0 where logs are base 2
adding logs is multiplying
log (x(4-x)) = 0
x(4-x) = 2^0 which is one
4x - x^2 = 1
x^2 -4x +1 = 0
solve for x
log base 2 (x) or log (2x)
I will assume base 2
then
log x + log (4-x) = 0 where logs are base 2
adding logs is multiplying
log (x(4-x)) = 0
x(4-x) = 2^0 which is one
4x - x^2 = 1
x^2 -4x +1 = 0
solve for x
2 answers