Self-base logarithms

Whenever we have a logarithm that’s in the form:

\log_a{a^x}

…the as essentially cancel out, and we just get x as the answer.

\log_a{a^x}=x

Evaluate \log_5{5^3}

• 5 to the power of what makes 5 to the power of 3?
• Well… 3.
• So our answer is 3!

Evaluate \log_{10}{10^{-2}}

• 10^x=10^{-2}
• x=-2
• Answer: -2

Evaluate \log_2{2^{\frac12}}

• 2^x=2^{\frac12}
• x=\frac12
• Answer: \frac12

Evaluate \log_3{3^0}

• 3^x=3^0
• x=0
• Answer: 0

Evaluate \log_7{7}

• 7^x=7^1
• x=1
• Answer: 1

Simplify \log_4{4^{2x+1}}

• 4^{y}=4^{2x+1}
• y=2x+1
• Answer: 2x+1

Natural logarithms of e

\ln x just means \log_e x, so \ln{e^x} is just \log_e{e^x}, which is just x.

\ln{e^x}=x

Evaluate \ln{e^4}

• =\log_e{e^4}
• e^x=e^4
• x=4
• Answer: 4

Evaluate \ln{e^{-3}}

• \ln{e^x}=x
• x=-3
• Answer: -3

Evaluate \ln{e^{\frac{x}{2}}}

• \ln{e^x}=x
• x=\frac{x}{2}
• Answer: \frac x2

Evaluate \ln{e^0}

• \ln{e^x}=x
• x=0
• Answer: 0

Evaluate \ln{e}

• \ln{e^x}=x
• e^x=e^1
• x=1
• Answer: 1

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