+0

33554432=x^5

0
557
4

33554432=x^5

Guest Jun 9, 2015

#4
+19638
+10

33554432=x^5   x = ?

$$\small{\text{ \rm{The~prime~ factorization~ of~} \mathbf{ 33554432 = 2^{25} } }} \\ \small{\text{ \rm{so~we~have~} \mathbf{ 2^{25} = x^5} }} \\ \small{\text{ \rm{or~} \mathbf{ 2^{5\cdot 5} = x^5} }} \\ \small{\text{ \rm{or~} \mathbf{ (2^5)^5 = (x)^5} \qquad | \qquad \rm {comparing~coefficient} }} \\ \small{\text{ \rm{so~} \mathbf{ (2^5) = x} }} \\ \small{\text{ \rm{or~} \mathbf{ 32 = x} }} \\$$

heureka  Jun 9, 2015
#1
+92781
+5

Raise each side to te power of 1/5

Melody  Jun 9, 2015
#2
+7348
+5

Hallo anonymous!

33554432=x^5

ln 33554432  = 5 * ln x

ln x = (ln 33554432) / 5 = 3,4657359028
x = e^3,4657359028

x = 32

Grüße von  :- )
asinus  Jun 9, 2015
#3
+92781
+5

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

$$\\33554432=x^5\\\\ x^5=33554432\\\\ (x^5)^{1/5}=33554432^{1/5}\\\\ x=33554432^{1/5}\\\\$$

$${{\mathtt{33\,554\,432}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right)} = {\mathtt{32}}$$

Melody  Jun 9, 2015
#4
+19638
+10
$$\small{\text{ \rm{The~prime~ factorization~ of~} \mathbf{ 33554432 = 2^{25} } }} \\ \small{\text{ \rm{so~we~have~} \mathbf{ 2^{25} = x^5} }} \\ \small{\text{ \rm{or~} \mathbf{ 2^{5\cdot 5} = x^5} }} \\ \small{\text{ \rm{or~} \mathbf{ (2^5)^5 = (x)^5} \qquad | \qquad \rm {comparing~coefficient} }} \\ \small{\text{ \rm{so~} \mathbf{ (2^5) = x} }} \\ \small{\text{ \rm{or~} \mathbf{ 32 = x} }} \\$$