Change the base of the following logarithms.
\log_4\left(62\right)
The formula to change the base of a logarithm is:
\log_a\left(b\right)=\dfrac{\log_m\left(b\right)}{\log_m\left(a\right)}
In our problem:
\log_4\left(62\right)=\dfrac{\log_2\left(62\right)}{\log_2\left(4\right)}
We know that:
\log_2\left(4\right)=2 because 2^{2}=4
We conclude that:
\log_4\left(62\right)=\dfrac{\log_2\left(62\right)}2
\log_9\left(73\right)
The formula to change the base of a logarithm is:
\log_a\left(b\right)=\dfrac{\log_m\left(b\right)}{\log_m\left(a\right)}
In our problem:
\log_9\left(73\right)=\dfrac{\log_3\left(73\right)}{\log_3\left(9\right)}
We know that:
\log_3\left(9\right)=2 because 3^{2}=9
We conclude that:
\log_9\left(73\right)=\dfrac{\log_3\left(73\right)}2
\log_2\left(\sqrt{e}\right)
The formula to change the base of a logarithm is:
\log_a\left(b\right)=\dfrac{\log_m\left(b\right)}{\log_m\left(a\right)}
In our problem:
\log_2\left(\sqrt{e}\right)=\dfrac{\ln\left(\sqrt{e}\right)}{\ln\left(2\right)}
We know that:
\ln\left(\sqrt{e}\right)=\dfrac{1}{2} because e^{\frac{1}{2}}=\sqrt{e}
We conclude that:
\log_2\left(\sqrt{e}\right)=\dfrac{1}{2\ln\left(2\right)}
\log_\dfrac{1}{100}\left(253\right)
The formula to change the base of a logarithm is:
\log_a\left(b\right)=\dfrac{\log_m\left(b\right)}{\log_m\left(a\right)}
In our problem:
\log_\dfrac{1}{100}\left(253\right)=\dfrac{\log\left(253\right)}{\log\left(\dfrac{1}{100}\right)}
We know that:
\log\left(\dfrac{1}{100}\right)=-2 because 10^{-2}=\dfrac{1}{100}
We conclude that:
\log_\dfrac{1}{100}\left(253\right)=-\dfrac{1}{2}\log\left(253\right)
\log_\sqrt[3]{5}\left(127\right)
The formula to change the base of a logarithm is:
\log_a\left(b\right)=\dfrac{\log_m\left(b\right)}{\log_m\left(a\right)}
In our problem:
\log_\sqrt[3]{5}\left(127\right)=\dfrac{\log_5\left(127\right)}{\log_5\left(\sqrt[3]{5}\right)}
We know that:
\log_5\left(\sqrt[3]{5}\right)=\dfrac{1}{3} because \sqrt[3]{5}=5^{\frac{1}{3}}
We conclude that:
\log_\sqrt[3]{5}\left(127\right)=3\log_5\left(127\right)
\log_8 \left(50\right)
The formula to change the base of a logarithm is:
\log_a\left(b\right)=\dfrac{\log_m\left(b\right)}{\log_m\left(a\right)}
In our problem:
\log_8\left(50\right)=\dfrac{\log_2\left(50\right)}{\log_2\left(8\right)}
We know that:
\log_2\left(8\right)=3 because 2^{3}=8
We conclude that:
\log_8\left(50\right)=\dfrac{\log_2\left(50\right)}{3}
\log_\dfrac{1}{\sqrt{e}} \left(12\right)
The formula to change the base of a logarithm is:
\log_a\left(b\right)=\dfrac{\log_m\left(b\right)}{\log_m\left(a\right)}
In our problem:
\log_\dfrac{1}{\sqrt{e}}\left(12\right)=\dfrac{\ln\left(12\right)}{\ln\left(\dfrac{1}{\sqrt{e}}\right)}
We know that:
\ln\left(\dfrac{1}{\sqrt{e}}\right)=-\dfrac{1}{2} because e^{-\frac{1}{2}}=\dfrac{1}{\sqrt{e}}
We conclude that:
\log_\dfrac{1}{\sqrt{e}}\left(12\right)=-2\ln\left(12\right)