Tough Logarithm Question

[sydney_kid]

So I've been trying this question for a while but it is really difficult as I don't know how to approach it? Any idea how to solve this?
I've tried substituting the x but I still am not getting anywhere, not because its impossible rather because I don't know how to. A point in the right direction would be well appreciated

 

[lex]

There is another solution, x=1.
I would change bases to e.g. the natural log.
[MATH](ax)^{\ln(x)/\ln(b)}=(bx)^{\ln(x)/\ln(a)}[/MATH]then write [MATH]ax[/MATH] as [MATH]e^{\ln(ax)}[/MATH] on lhs, [MATH]bx[/MATH] as [MATH] e^{\ln(bx)}[/MATH] on rhs
[MATH](e^{\ln(ax)})^{\ln(x)/\ln(b)} = (e^{\ln(bx)})^{\ln(x)/\ln(a)}\\ e^{\ln(x)\ln(ax)/\ln(b)}=e^{\ln(x)\ln(bx)/\ln(a)}\\ \ln(x)\ln(ax)/\ln(b)=\ln(x)\ln(bx)/\ln(a) \hspace2ex \text{(since }e^X \text{ is 1-1)}\\ \ln(x)=0\text{ or } \ln(ax)/\ln(b)=\ln(bx)/\ln(a) \\ \text{i.e. }x=1 \text{ or }\ln(ax)/\ln(b)=\ln(bx)/\ln(a)[/MATH]Then use rules of logs to isolate [MATH]\ln(x)[/MATH] and you should get what you want.

...
Finally solutions:
x=1 or x=1/ab

 

[greg1313]

x=1 or x=1/ab

x = 1 \(\displaystyle \ \) or \(\displaystyle \ \) x = 1/(ab)