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$\log_{ \left(2 \sqrt{ 2 } \right) } {\left( \sqrt[ 4 ]{ 32 } \right)}$
$\dfrac { 5 } { 6 }$
Calculate the value
$\log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ \color{#FF6800}{ 4 } ]{ \color{#FF6800}{ 32 } } \right) }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$\log _{ 2 \sqrt{ 2 } } { \left( \color{#FF6800}{ 2 } \sqrt[ \color{#FF6800}{ 4 } ]{ \color{#FF6800}{ 2 } } \right) }$
$\log _{ \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 2 } } } { \left( \color{#FF6800}{ 2 } \sqrt[ \color{#FF6800}{ 4 } ]{ \color{#FF6800}{ 2 } } \right) }$
$ $ Simplify the expression using $ \log_{a}{x\times y}=\log_{a}{x}+\log_{a}{y}$
$\log _{ \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 2 } } } { \left( \color{#FF6800}{ 2 } \right) } \color{#FF6800}{ + } \log _{ \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 2 } } } { \left( \sqrt[ \color{#FF6800}{ 4 } ]{ \color{#FF6800}{ 2 } } \right) }$
$\log _{ \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 2 } } } { \left( \color{#FF6800}{ 2 } \right) } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$ $ Rationalize the base of the logarithm $ $
$\color{#FF6800}{ 2 } \log _{ \color{#FF6800}{ 8 } } { \left( \color{#FF6800}{ 2 } \right) } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$2 \log _{ 8 } { \left( \color{#FF6800}{ 2 } \right) } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$ $ Write the number in exponential form with base $ 2$
$2 \log _{ 2 ^ { 3 } } { \left( \color{#FF6800}{ 2 } \right) } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$2 \log _{ \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 3 } } } { \left( \color{#FF6800}{ 2 } \right) } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$ $ Simplify the expression using $ \log_{a^{y}}{a}=\dfrac{1}{y}\log_{a}{a}$
$2 \times \color{#FF6800}{ \dfrac { 1 } { 3 } } \log _{ \color{#FF6800}{ 2 } } { \left( \color{#FF6800}{ 2 } \right) } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$2 \times \dfrac { 1 } { 3 } \log _{ \color{#FF6800}{ 2 } } { \left( \color{#FF6800}{ 2 } \right) } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$ $ The logarithm is equal to 1 if a base is same as an antilogarithm $ $
$2 \times \dfrac { 1 } { 3 } \times \color{#FF6800}{ 1 } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$2 \times \dfrac { 1 } { 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$ $ Multiplying any number by 1 does not change the value $ $
$2 \times \dfrac { 1 } { 3 } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$\color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$ $ Calculate the product of rational numbers $ $
$\color{#FF6800}{ \dfrac { 2 } { 3 } } + \log _{ 2 \sqrt{ 2 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$\dfrac { 2 } { 3 } + \log _{ \color{#FF6800}{ 2 } \sqrt{ \color{#FF6800}{ 2 } } } { \left( \sqrt[ \color{#FF6800}{ 4 } ]{ \color{#FF6800}{ 2 } } \right) }$
$ $ Rationalize the base of the logarithm $ $
$\dfrac { 2 } { 3 } + \color{#FF6800}{ 2 } \log _{ \color{#FF6800}{ 8 } } { \left( \sqrt[ \color{#FF6800}{ 4 } ]{ \color{#FF6800}{ 2 } } \right) }$
$\dfrac { 2 } { 3 } + 2 \log _{ \color{#FF6800}{ 8 } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$ $ Write the number in exponential form with base $ 2$
$\dfrac { 2 } { 3 } + 2 \log _{ \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 3 } } } { \left( \sqrt[ 4 ]{ 2 } \right) }$
$\dfrac { 2 } { 3 } + 2 \log _{ \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ 3 } } } { \left( \sqrt[ \color{#FF6800}{ 4 } ]{ \color{#FF6800}{ 2 } } \right) }$
$ $ Simplify the expression using $ \log_{a^{y}}{b}=\dfrac{1}{y}\times\log_{a}{b}$
$\dfrac { 2 } { 3 } + 2 \times \color{#FF6800}{ \dfrac { 1 } { 3 } } \log _{ \color{#FF6800}{ 2 } } { \left( \sqrt[ \color{#FF6800}{ 4 } ]{ \color{#FF6800}{ 2 } } \right) }$
$\dfrac { 2 } { 3 } + 2 \times \dfrac { 1 } { 3 } \log _{ 2 } { \left( \sqrt[ \color{#FF6800}{ 4 } ]{ \color{#FF6800}{ 2 } } \right) }$
$ $ Convert the square root of the antilogarithm number of the logarithm to the power $ $
$\dfrac { 2 } { 3 } + 2 \times \dfrac { 1 } { 3 } \log _{ 2 } { \left( \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ \frac { 1 } { 4 } } } \right) }$
$\dfrac { 2 } { 3 } + 2 \times \dfrac { 1 } { 3 } \log _{ \color{#FF6800}{ 2 } } { \left( \color{#FF6800}{ 2 } ^ { \color{#FF6800}{ \frac { 1 } { 4 } } } \right) }$
$ $ Simplify the expression using $ \log_{a}{a^{x}}=x\times\log_{a}{a}$
$\dfrac { 2 } { 3 } + 2 \times \dfrac { 1 } { 3 } \times \color{#FF6800}{ \dfrac { 1 } { 4 } } \log _{ \color{#FF6800}{ 2 } } { \left( \color{#FF6800}{ 2 } \right) }$
$\dfrac { 2 } { 3 } + 2 \times \dfrac { 1 } { 3 } \times \dfrac { 1 } { 4 } \log _{ \color{#FF6800}{ 2 } } { \left( \color{#FF6800}{ 2 } \right) }$
$ $ The logarithm is equal to 1 if a base is same as an antilogarithm $ $
$\dfrac { 2 } { 3 } + 2 \times \dfrac { 1 } { 3 } \times \dfrac { 1 } { 4 } \times \color{#FF6800}{ 1 }$
$\dfrac { 2 } { 3 } + 2 \times \dfrac { 1 } { 3 } \times \dfrac { 1 } { 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 }$
$ $ Multiplying any number by 1 does not change the value $ $
$\dfrac { 2 } { 3 } + 2 \times \dfrac { 1 } { 3 } \times \dfrac { 1 } { 4 }$
$\dfrac { 2 } { 3 } + \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 4 } }$
$ $ Calculate the product of rational numbers $ $
$\dfrac { 2 } { 3 } + \color{#FF6800}{ \dfrac { 1 } { 6 } }$
$\color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 6 } }$
$ $ Find the sum of the fractions $ $
$\color{#FF6800}{ \dfrac { 5 } { 6 } }$
Solution search results
search-thumbnail-If the sum of two consecutive 
numbers is $45$ and one number is $X$ 
.This statement in the form of 
equation $1s:$ 
$\left(1$ Point) $\right)$ 
$○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ 
$○sx+1ef\left(x+2$ $r1gnt\right)=145s$ 
$sx+1x=45s$
7th-9th grade
Algebra
search-thumbnail-$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ 
$s1S$ 
$S-1S$ 
$s2S$ 
$s50s$
7th-9th grade
Other
search-thumbnail-The rationalizing factor of \sqrt{23} is 
$°$ $Options^{°}$ $0$ 
A 24 
23 
C \sqrt{23} 
D None of these
7th-9th grade
Other
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