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Using Properties of Logarithms

Using Properties of Logarithms 1.16 - Solution

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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.

log5(x)=2\log_5(x)=2
5LHS=5RHS5^{\text{LHS}}=5^{\text{RHS}}
5log5(x)=525^{\log_5(x)}=5^2
5log5(m)=m{5}^{\log_{5}(m)}=m
x=52x=5^2
x=25x=25
b

Here we can use the same method.

log4(x)=-1\log_4{(x)}=\text{-}1
4LHS=4RHS4^{\text{LHS}}=4^{\text{RHS}}
4log4(x)=4-14^{\log_4{(x)}}=4^{\text{-}1}
4log4(m)=m{4}^{\log_{4}(m)}=m
x=4-1x=4^{\text{-}1}
x=14x=\dfrac{1}{4}
x=0.25x=0.25
c

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

5log2(x)+80=1005\log_2{(x)}+80=100
5log2(x)=205\log_2{(x)}=20
log2(x)=4\log_2{(x)}=4
2LHS=2RHS2^{\text{LHS}}=2^{\text{RHS}}
2log2(x)=242^{\log_2{(x)}}=2^4
2log2(m)=m{2}^{\log_{2}(m)}=m
x=24x=2^4
x=16x=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.

log7(2x)2=1\dfrac{\log_7{(2x)}}{2}=1
log7(2x)=2\log_7(2x)=2
7LHS=7RHS7^{\text{LHS}}=7^{\text{RHS}}
7log7(2x)=727^{\log_7(2x)}=7^2
7log7(m)=m{7}^{\log_{7}(m)}=m
2x=722x=7^2
2x=492x=49
x=24.5x=24.5