Find x - logarithmic equation?

Question

Solve log_x(5) - log_5(x) = log_5(1\fifth root of x)

Answer 1

I'm assuming that the RHS of the equation reads: logarithm to base 5

of 1 over the fifth root of x, log_5 (1/⁵√x) or log_5 [x^(-1/5)].



log_x (5) - log_5 (x) = log_5 [x^(-1/5)].

Replacing log_x (5) with its equivalent 1/log_5(x)

1/log_5 (x) - log_5 (x) = (-1/5) log_5 (x).

Let log_5(x) = y,

1/y - y = (-1/5)y

1/y = y - (1/5)y

1/y = 4/5y

y² = 5/4

y = ±√(5)/2

Hence

log_5 (x) = ±√(5)/2

x = 5^(±√(5)/2) = √(5)^(±√(5))

x = √(5)^√(5) or x = 1/√(5)^√(5)

Answer 2

log_x(5) -1/( log_x(5))=- (1/5) log_x(5)

considering y=log_x(5)

y- 1/y=-y/5

y^2-1=(y^2)/5

4y^2==5

y^2=5/4

y=±(√5)/2

log_5(x)=√5/2

5=x^(√5/2)

x=5^(2/√5)

Answer 3

log[x] 5 - log[5] x = log[5] √x /5



1/log[5] x - log[5] x = log[5] √x - log[5] 5



1/log[5] x - log[5] x = 1/2 log[5] x - 1



Let log[5] x = a



1/a - a = a/2 - 1

Multiply each term by 2a

2 - 2a^2 = a^2 - 2a

3a^2 - 2a - 2 = 0...not factorisable!



log_5(1\fifth root of x)?? Creating confusion.

Hopefully you get the idea