Question

4(3n-2)^(-3/2)+1=347/343

Answer 1

Greetings and Happy Holidays!

Solve for n.
$4(3n-2)^{\frac{-3}{2}}+1= \frac{347}{343}$

Add -1 to both sides.
$(4(3n-2)^{\frac{-3}{2}}+1)+(-1)=(\frac{347}{343})+(-1)$

$4(3n-2)^{\frac{-3}{2}}=\frac{347}{343}-1$

$4(3n-2)^{\frac{-3}{2}}=\frac{347}{343}-\frac{343}{343}$

$4(3n-2)^{\frac{-3}{2}}=\frac{4}{343}$

Divide both sides by 4
$\frac{4(3n-2)^{\frac{-3}{2}}}{4}=\frac{\frac{4}{343}}{4}$

$\frac{4(3n-2)^{\frac{-3}{2}}}{4}=\frac{4}{343}(\frac{1}{4})$

$\frac{4(3n-2)^{\frac{-3}{2}}}{4}=\frac{4}{1372}$

$\frac{4(3n-2)^{\frac{-3}{2}}}{4}=\frac{1}{343}$

$(3n-2)^{\frac{-3}{2}}=\frac{1}{343}$

Solve the Exponent.
$((3n-2)^{\frac{-3}{2}})^{\frac{2}{-3}} =\frac{1}{343}^{\frac{2}{-3}}$

$3n-2=49$

Add 2 to both sides.
$(3n-2)+2=(49)+2$

$3n=51$

Divide both sides by 3.
$\frac{3n}{3}= \frac{51}{3}$

$n=17$

The Answer Is:
$\left[\begin{array}{ccc}n=17\end{array}\right]$

Hope this helped! 
-Benjamin