_{2}x + log

_{x}2=2.5

solve equation log_{2}x + log_{x}2=2.5

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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

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