Solve the equation #25log_2x=log_x2# ?

#25log_2x=log_x2#
I have gotten to #logx^25=1/(logx)# but I am not sure how to proceed from here. I am supposed to solve this without the use of a calculator, and the answer is given as #x=2^((+/-)1/5)#. How should I go about solving this?

2 Answers
Apr 22, 2018

I tried this:

Explanation:

I would try changing the base of the second log as:

#log_x2=log_2(2)/log_2(x)=1/log_2(x)#

so we get:

#25log_2x=1/log_2(x)#

rearrange:
#25log_2x*log_2x=1#
#25[log_2x]^2=1#
#[log_2x]^2=1/25#
take the square root of both sides to get rid of the square:
#log_2x=+-1/5#
apply the definition of log:
#x=2^(+-1/5)#

Apr 22, 2018

#x=1/2^(1/5)#, #x=2^(1/5)#

Explanation:

#25log_2(x)=log_x(2)#
#log_2(x^(25))=log_x(2)#
#log(x^(25))/log(2)=log(2)/log(x)#
#25log(x)*log(x)=log(2)*log(2)#
#25log^2(x)=log^2(2)#
#5log(x)=+-log(2)#
#log(x^5)=+-log(2)#

#log(x^5)=log(2^(-1))=log(1/2)#
#x^5=1/2#
#x=1/2^(1/5)#

#log(x^5)=log(2)#
#x^5=2#
#x=2^(1/5)#