solve the equation log2 x + logx 2 =2.5

Question

solve equation log₂x + log_x2=2.5

Answer 1

First change the base on log_x2 using the change of base identity:

log_x2=log₂2/log₂x=1/log₂x

Then your equation becomes

log₂x + 1/log₂x =2.5

Let u=log₂x 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=log₂x,  x=4.

For u=1/2=log₂x, x=√2

Therefore, the two solutions are x=4 and x=√2

 

Check: log₂4 + log₄2 = 2+1/2=2.5

log₂√2 + log _√22 = 1/2 + 2 =2.5

Answer 2

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

 

Answer 3

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

 

log₂x + log_x2 = 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

 

ln²(x) + ln²(2) = 2.5 ln(x)ln(2)

 

ln(2) = 0.693

2.5ln(2) = 1.733

ln²(2) = 0.480

 

 

let y = ln(x) your eqn will become

 

y² -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(mⁿ) = n · log_b(m)

 

in our case

 

verify x =4

 

ln(4)/ln(2) + ln(2)/ln(4) = 2.5

 

ln(2²)/ln(2) + ln(2)/ln(2²) = 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