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# Using Properties of Logarithms

## Using Properties of Logarithms 1.16 - Solution

a

The inverse properties of logarithms state that a logarithm and a power with the same base undo each other. Let's use this to solve the equation.

$\log_5(x)=2$
$5^{\text{LHS}}=5^{\text{RHS}}$
$5^{\log_5(x)}=5^2$
${5}^{\log_{5}(m)}=m$
$x=5^2$
$x=25$
b

Here we can use the same method.

$\log_4{(x)}=\text{-}1$
$4^{\text{LHS}}=4^{\text{RHS}}$
$4^{\log_4{(x)}}=4^{\text{-}1}$
${4}^{\log_{4}(m)}=m$
$x=4^{\text{-}1}$
$x=\dfrac{1}{4}$
$x=0.25$
c

Here we first need to isolate the logarithm on the left-hand side.

$5\log_2{(x)}+80=100$
$5\log_2{(x)}=20$
$\log_2{(x)}=4$
$2^{\text{LHS}}=2^{\text{RHS}}$
$2^{\log_2{(x)}}=2^4$
${2}^{\log_{2}(m)}=m$
$x=2^4$
$x=16$
d

Just like in the previous exercise we first need to isolate the logarithm before we can undo it using the inverse properties of logarithms.

$\dfrac{\log_7{(2x)}}{2}=1$
$\log_7(2x)=2$
$7^{\text{LHS}}=7^{\text{RHS}}$
$7^{\log_7(2x)}=7^2$
${7}^{\log_{7}(m)}=m$
$2x=7^2$
$2x=49$
$x=24.5$