Using Properties of Logarithms

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

CalcPowCalculate power

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}

a^{\text{-} 1}=\dfrac{1}{a}

x=\dfrac{1}{4}

WriteDecWrite as a decimal

x=0.25

c

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

5\log_2{(x)}+80=100

\text{LHS}-80=\text{RHS}-80

5\log_2{(x)}=20

\left.\text{LHS}\middle/5\right.=\left.\text{RHS}\middle/5\right.

\log_2{(x)}=4

2^{\text{LHS}}=2^{\text{RHS}}

2^{\log_2{(x)}}=2^4

{2}^{\log_{2}(m)}=m

x=2^4

CalcPowCalculate power

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

\text{LHS} \cdot 2=\text{RHS}\cdot 2

\log_7(2x)=2

7^{\text{LHS}}=7^{\text{RHS}}

7^{\log_7(2x)}=7^2

{7}^{\log_{7}(m)}=m

2x=7^2

CalcPowCalculate power

2x=49

\left.\text{LHS}\middle/2\right.=\left.\text{RHS}\middle/2\right.

x=24.5

Using Properties of Logarithms 1.16 - Solution