Exponential inequality solution I am trying to solve 
$$  \displaystyle (2^{(3x-1)/(x-1)})^{1/3} < 8^{((x-3)/(3x-7))}$$
Here's how I proceeded.
$LHS = 2^{((3x-1)/(3(x-1))}  , RHS = 2^{((3*(x-3))/(3x-7))}$
hence inequating the exponents ,
$$\frac {(3x-1)}{(3(x-1))} < 3\frac {(x-3)}{(3x-7)}$$
$$ \implies  (3x-1)(3x-7) < 9(x-1)(x-3)$$
$$\implies    9x^2-24x+7 < 9x^2-36x+27$$
cancelling terms
$$12x <20$$
$$=> x < 5/3 $$
but the book states the solution as $(-\infty , 1)$ U $(5/3,7/3)$
Not sure how this has been arrived at , can someone help me understand ?
Thanks in advance.
Madavan.
 A: Be carefull when you multiply your inequality
$$\frac {(3x-1)}{(3(x-1))} < 3\frac {(x-3)}{(3x-7)}$$
Because $3x-7$ or $3(x-1)$ may be negative and change the order of the ineqaulity
You need to study the sign of $3(x-1)(3x-7)$...
A)$ \,3(x-1)(3x-7) $ is positive
$$\implies  (3x-1)(3x-7) < 9(x-1)(x-3)$$
$$ \implies x<5/3$$
First case $3(x-1)>0$ and $3x-7>0 \implies x> \frac 73$
With the constraint $x>7/3$ there are no solution.
Second case $3(x-1)<0$ and $3x-7<0$
$$\implies x<1$$
with this constraint ($x<1$) you get a part of the result of the book $ \implies x \in [-\infty,1]$. 
Now you have to study the case when $3(x-1)(3x-7)$ is negative to get the other part of the solution of the book...
B)$ \,3(x-1)(3x-7) $ is negative
$$  (3x-1)(3x-7) > 9(x-1)(x-3)$$
$$ \implies  x>\frac 5 3$$
First case
$3(x-1)<0 \quad ; 3x-7>0 \implies x<1, x>7/3$ Impossible ..
Second case
$3(x-1)>0 \quad ; 3x-7<0 \implies x>1, x<7/3$ .
With the constraint you finally get $x \in (5/3,7/3)$
And the complete solution is $(-\infty,1)$ U $(5/3,7/3)$
