Convert between a difference of logarithms and a quotient Precalculus

The formula for a difference of logarithms with the same base is:

\log_a\left(x\right)-\log_a\left(y\right)=\log_a\left(\dfrac{x}{y}\right)

Applying this formula to our problem:

\log_2\left(3x\right)-\log_2\left(6x\right)=\log_2\left(\dfrac{3x}{6x}\right)=\log_2\left(\dfrac{3}{6}\right)=\log_2\left(\dfrac{1}{2}\right)

When calculating \log_a\left(b\right), we are looking for a real number c so that:

a^{c}=b

In our problem:

2^{c}=\dfrac{1}{2}

Since:

\dfrac{1}{2}=2^{-1}

We conclude that:

\log_2\left(\dfrac{1}{2}\right)=-1

The formula for a difference of logarithms with the same base is:

\log_a\left(x\right)-\log_a\left(y\right)=\log_a\left(\dfrac{x}{y}\right)

Applying this formula to our problem:

\log_3\left(9x^{2}\right)-\log_3\left(3x^{2}\right)=\log_3\left(\dfrac{9x^{2}}{3x^{2}}\right)=\log_3\left(3\right)

When calculating \log_a\left(b\right), we are looking for a real number c so that:

a^{c}=b

In our problem:

3^{c}=3

Since:

3=3^{1}

We conclude that:

\log_3\left(3\right)=1

The formula for a difference of logarithms with the same base is:

\log_a\left(x\right)-\log_a\left(y\right)=\log_a\left(\dfrac{x}{y}\right)

Applying this formula to our problem:

\log_5\left(\sqrt[3]{625 m^{2}}\right)-\log_5\left(\sqrt[3]{25 m^{2}}\right)=\log_5\left(\dfrac{\sqrt[3]{625 m^{2}}}{\sqrt[3]{25 m^{2}}}\right)=\log_5\left(\sqrt[3]{25}\right)

When calculating \log_a\left(b\right), we are looking for a real number c so that:

a^{c}=b

In our problem:

5^{c}=\sqrt[3]{25}

Since:

\sqrt[3]{25}=5^{\frac{2}{3}}

We conclude that:

\log_5\left(\sqrt[3]{25}\right)=\dfrac{2}{3}

The formula for a difference of logarithms with the same base is:

\log_a\left(x\right)-\log_a\left(y\right)=\log_a\left(\dfrac{x}{y}\right)

Applying this formula to our problem:

\ln\left(x^{2}\right)-\ln\left(ex^{2}\right)=\ln\left(\dfrac{x^{2}}{ex^{2}}\right)=\ln\left(\dfrac{1}{e}\right)

When calculating \log_a\left(b\right), we are looking for a real number c so that:

a^{c}=b

In our problem:

e^{c}=\dfrac{1}{e}

Since:

\dfrac{1}{e}=e^{-1}

We conclude that:

\ln\left(\dfrac{1}{e}\right)=-1

The formula for a difference of logarithms with the same base is:

\log_a\left(\dfrac{x}{y}\right)=\log_a\left(x\right)-\log_a\left(y\right)

Applying this formula to our problem:

\log_2\left(\dfrac{8}{7}\right)=\log_2\left(8\right)-\log_2\left(7\right)

When calculating \log_a\left(b\right), we are looking for a real number c so that:

a^{c}=b

In our problem, when calculating \log_2\left(8\right) :

2^{c}=8

Since:

8=2^{3}

We conclude that:

\log_2\left(8\right)=3

The formula for a difference of logarithms with the same base is:

\log_a\left(\dfrac{x}{y}\right)=\log_a\left(x\right)-\log_a\left(y\right)

Applying this formula to our problem:

\ln\left(\dfrac{3}{e^{2}}\right)=\ln\left(3\right)-\ln\left(e^{2}\right)

When calculating \log_a\left(b\right), we are looking for a real number c so that:

a^{c}=b

In our problem, when calculating \ln\left(e^{2}\right) :

e^{c}=e^{2}

Since:

e^{2}=e^{2}

We conclude that:

\ln\left({e^{2}}\right)=2

The formula for a difference of logarithms with the same base is:

\log_a\left(\dfrac{x}{y}\right)=\log_a\left(x\right)-\log_a\left(y\right)

Applying this formula to our problem:

\log\left(\dfrac{\sqrt[4]{100}}{7}\right)=\log\left(\sqrt[4]{100}\right)-\log\left(7\right)

When calculating \log_a\left(b\right), we are looking for a real number c so that:

a^{c}=b

In our problem, when calculating \log\left(\sqrt[4]{100}\right) :

10^{c}=\sqrt[4]{100}

Since:

\sqrt[4]{100}=10^{\frac{2}{4}}=10^{\frac{1}{2}}

We conclude that:

\log\left(\sqrt[4]{100}\right)=\dfrac{1}{2}