2[log_4(x) + log_x(4)] = 5 ?

Question

Find the possible values of x?

Answer is 2 and 16.

Please show it in detail..

thanx a million..

Answer 1

Divide both sides by 2:
log_4(x) + log_x(4) = 5/2

Use the change of base rule to make the second have a base of 4: log_a(b) = log(b) / log(a)

log_4(x) + log_4(4) / log_4(x) = 5/2

log_4(4) becomes 1:
log_4(x) + 1 / log_4(x) = 5/2

Multiply both sides by log_4(x) to clear the denominator:
[ log_4(x) ]² + 1 = 5/2 log_4(x)

Get everything on one side:
[ log_4(x) ]² - 5/2 log_4(x) + 1 = 0

Now substitute k = log_4(x)

k² - 5/2k + 1 = 0

Multiply by 2:
2k² - 5k + 2 = 0

This factors as:
(2k - 1)(k - 2) = 0

So k = 2 or k = ½

Now replace back:
log_4(x) = 2
or
log_4(x) = ½

Raise both sides upon the power 4:
x = 4^2 = 16
x = 4^½ = 2

Answer 2

Easy, first change the bases of the logs to 4

Then you'll get:

2[log_4(x) + log_4(4)/log_4(x)] = 5

Then let log_4(x) = y

2[y + 1/y] = 5

2y^2 + 2 = 5y

2y^2 - 5y + 2 = 0

Then by using Bhaskara you'll get

y = 1/2

y' = 2

Then again log_4(x) = y

We'll have 2 values:

With y -> 4^y = x -> 4^1/2= x -> x = 2

With y' ->4^y' = x' -> 4^2 = x' -> x' = 16

Cya

Answer 3

change base on log_x(4)

becomes
log_4(4) / log_4(x)
= 1 / log_4(x)

Now you have:
2[log_4(x) + 1/log_4(x)] = 5

let y = log_4(x)

2(y + 1/y) = 5
2[(y^2 + 1) / y] = 5
2(y^2 + 1) = 5y
2y^2 + 2 = 5y
2y^2 - 5y + 2 = 0
(2y - 1)(y - 2) = 0
so...
y = 1/2 and 2

which means
log_4(x) = 1/2 and 2

log_4(x) = 1/2
x = 4^(1/2) = sqrt 4 = 2

log_4(x) = 2
x = 4^2 = 16