Solve the logarithmic equation $\log_{32}\left(x\right)=\frac{1}{5}$

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Final answer to the problem

$x=2$
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Change the logarithm to base $10$ applying the change of base formula for logarithms: $\log_b(a)=\frac{\log_{10}(a)}{\log_{10}(b)}$. Since $\log_{10}(b)=\log(b)$, we don't need to write the $10$ as base

$\frac{\log \left(x\right)}{\log \left(32\right)}=\frac{1}{5}$

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$\frac{\log \left(x\right)}{\log \left(32\right)}=\frac{1}{5}$

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Learn how to solve logarithmic equations problems step by step online. Solve the logarithmic equation log32(x)=1/5. Change the logarithm to base 10 applying the change of base formula for logarithms: \log_b(a)=\frac{\log_{10}(a)}{\log_{10}(b)}. Since \log_{10}(b)=\log(b), we don't need to write the 10 as base. Apply fraction cross-multiplication. Apply the formula: a\log_{b}\left(x\right)=\log_{b}\left(x^a\right), where a=5 and b=10. For two logarithms of the same base to be equal, their arguments must be equal. In other words, if \log(a)=\log(b) then a must equal b.

Final answer to the problem

$x=2$

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Function Plot

Plotting: $\log_{32}\left(x\right)-\frac{1}{5}$

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7
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9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Logarithmic Equations

Are those equations in which the unknown variable appears within a logarithm.

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