What is the solution for $x^{\frac {10x}{3} }=\frac {243} {32}$? I understand that $243$ is $3^5$ and $32$ is $2^5$, but I don't know where to go from there after converting it to logarithm.
 A: $$x^\frac{10x}{3}=\frac{3^5}{2^5}$$
Power both sides by $1/5$, so
$$x^\frac{2x}{3}=\frac{3}{2}$$
Taking logarithms
$$\tfrac{2x}{3}\log(x)=\log(\tfrac{3}{2})$$
$$x\log(x)=\tfrac{3}{2}\log(\tfrac{3}{2})$$
So $x=3/2$
A: You are nearly there 
$x^{\frac {10x}{3} }=(\frac {3} {2})^5$?
Well, now just see if x = ($\frac {3} {2}$) satisfies the answer. It does. So x = ($\frac {3} {2}$)
A: take logs on either side
$\frac{10}{3} x ln x = 5 \ln \frac{3}{2}$. Now substituting (3/2) satisfies above equation. So x = (3/2)
A: You have : x10x/3 = 243/32 
As you mentioned, 243 is 35 and 32 is 25
The equation can be re-written as:
x10x/3=(3/2)5
Applying logarithms on both sides, and using the power rule to drop down the exponents.
10x/3 log(x) = 5 log(3/2)
Moving 10/3 to the RHS.
x log(x)= 5*3/10 log(3/2)
x log(x) = 15/10 log(3/2)
15/10 is 3/2,
Therefore,
x log(x) = 3/2 log(3/2)
You can now determine that x is 3/2 or 1.5
A: As you said, $\frac{243}{32}=(\frac{3}{2})^5$. So why not try $x=\frac{3}{2}$? Indeed, we have $\frac{10\cdot \frac{3}{2}}{3}=5$ which makes $x=\frac{3}{2}$ a solution. We will now Show that there are no other solutions.
Obviously, we cannot have $0<x<1$ because that would imply $x^{\frac{10x}{3}}<1$. Furthermore, if we had $x<-1$, it would follow that $x^{\frac{10x}{3}}<1$ as well (if the term even was defined). And then, $x$ can't be between $0$ and $-1$ because that would make the exponent no whole number - but powers with this kind of exponent are only defined for positive reals as the base. So we have $x>1$. Thus, we consider the function $x^{\frac{10x}{3}}$ on the interval $(1, \infty)$. The derivative of this function is $$(x^{\frac{10x}{3}})'=(e^{ln(x)\cdot \frac{10x}{3}})'=e^{ln(x)\cdot \frac{10x}{3}}\cdot (ln(x)\cdot \frac{10}{3})\cdot (\frac{1}{x}\cdot \frac{10x}{3})=\frac{100}{9}\cdot e^{ln(x)\cdot \frac{10x}{3}}\cdot ln(x)>0$$
for all $x\in (1, \infty)$. So our function is strictly monotonically increasing on the interval $(1, \infty)$ and thus there can only be this one solution we discovered above.
