33554432=x^5 x = ?
$$\small{\text{$
\rm{The~prime~ factorization~ of~} \mathbf{ 33554432 = 2^{25} }
$}} \\
\small{\text{$
\rm{so~we~have~} \mathbf{ 2^{25} = x^5}
$}} \\
\small{\text{$
\rm{or~} \mathbf{ 2^{5\cdot 5} = x^5}
$}} \\
\small{\text{$
\rm{or~} \mathbf{ (2^5)^5 = (x)^5} \qquad | \qquad \rm {comparing~coefficient}
$}} \\
\small{\text{$
\rm{so~} \mathbf{ (2^5) = x}
$}} \\
\small{\text{$
\rm{or~} \mathbf{ 32 = x}
$}} \\$$
Hallo anonymous!
33554432=x^5
ln 33554432 = 5 * ln x
Thanks asinus
You have done it the really long way. You only need to use logs when you are finding a power.
The best way to do this problem is to just raise both sides to the power of 1/5
$$\\33554432=x^5\\\\
x^5=33554432\\\\
(x^5)^{1/5}=33554432^{1/5}\\\\
x=33554432^{1/5}\\\\$$
$${{\mathtt{33\,554\,432}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right)} = {\mathtt{32}}$$
33554432=x^5 x = ?
$$\small{\text{$
\rm{The~prime~ factorization~ of~} \mathbf{ 33554432 = 2^{25} }
$}} \\
\small{\text{$
\rm{so~we~have~} \mathbf{ 2^{25} = x^5}
$}} \\
\small{\text{$
\rm{or~} \mathbf{ 2^{5\cdot 5} = x^5}
$}} \\
\small{\text{$
\rm{or~} \mathbf{ (2^5)^5 = (x)^5} \qquad | \qquad \rm {comparing~coefficient}
$}} \\
\small{\text{$
\rm{so~} \mathbf{ (2^5) = x}
$}} \\
\small{\text{$
\rm{or~} \mathbf{ 32 = x}
$}} \\$$