solve equation log_{2}x + log_{x}2=2.5
solve the equation log2 x + logx 2 =2.5
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3 Answers
First change the base on log_{x}2 using the change of base identity:
log_{x}2=log_{2}2/log_{2}x=1/log_{2}x
Then your equation becomes
log_{2}x + 1/log_{2}x =2.5
Let u=log_{2}x and simplify:
u+1/u=2.5
u²-2.5u+1=0
Use the quadratic formula to solve this quadratic equation, get u = 2 or 1/2.
For u=2=log_{2}x, x=4.
For u=1/2=log_{2}x, x=√2
Therefore, the two solutions are x=4 and x=√2
Check: log_{2}4 + log_{4}2 = 2+1/2=2.5
log_{2}√2 + log _{√2}2 = 1/2 + 2 =2.5
Use the change of base formula,
log2 x + 1/log2 x = 2.5
Let u = log2 x,
u + 1/u = 2.5
Multiply both sides by 2u, and collect all terms in one side,
2u^2 - 5u + 2 = (2u-1)(u-2) = 0
u = 1/2, x = sqrt(2)
or
u = 2, x = 2^2 = 4
Answer: x = sqrt(2), and x = 4
if you remember that if you were going to change base of a log
log_{b}(x) = log_{d}(x) / log_{d}(b)
if we used this principle with your eqn
log_{2}x + log_{x}2 = 2.5
we could get the eqn into on base
ln(x)/ln(2) + ln(2)/ln(x) = 2.5
multiply by ln(2)ln(x) to remove denominator
ln^{2}(x) + ln^{2}(2) = 2.5 ln(x)ln(2)
ln(2) = 0.693
2.5ln(2) = 1.733
ln^{2}(2) = 0.480
let y = ln(x) your eqn will become
y^{2} -1.733y + 0.480 = 0
Solving for y using quadratic eqn
you get
y = 1.3869 and y = 0.3461
but we need to solve for x
x = e^{y}
x = 4 and x = 1.4135 = sqrt(2)
to verify if you remember
log_{b}(m^{n}) = n · log_{b}(m)
in our case
verify x =4
ln(4)/ln(2) + ln(2)/ln(4) = 2.5
ln(2^{2})/ln(2) + ln(2)/ln(2^{2}) = 2.5
2ln(2)/ln(2) +ln(2)/(2ln(2)) = 2 + 1/2 = 2.5
verify x = sqrt(2) = 2^{1/2}
ln(2)/(ln(sqrt(2))) + ln(sqrt(2))/ln(2) =
ln(2)/(1/2ln(2) + 1/2ln(2)/ln(2)
1/(1/2) + 1/2 = 2 + 1/2 = 2.5