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The Laws of Logs Simplified Revision Notes

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The Laws of Logs

Logarithms are the inverse of exponentials. If ay=xa^y = x, then loga(x)=y\log_a(x) = y. This means logarithms answer the question: "To what power must the base aa be raised to get xx?"

Recap:

  1. Base: The number aa is the base of the logarithm, and it must be a>0a > 0 and a1a \neq 1.
  2. Argument: The value xx inside the logarithm (loga(x)\log_a(x)) is called the argument and must satisfy x>0x > 0.

The following are the key laws of logarithms, which help simplify expressions and solve equations involving logs:

  1. Product Rule: loga(xy)=loga(x)+loga(y)\log_a(xy) = \log_a(x) + \log_a(y)

This means the logarithm of a product equals the sum of the logarithms of the factors.

  1. Quotient Rule: loga(xy)=loga(x)loga(y)\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)

The logarithm of a quotient equals the difference of the logarithms of the numerator and denominator.

  1. Power Rule: loga(xq)=qloga(x)\log_a(x^q) = q \cdot \log_a(x)

If the argument has an exponent, you can bring the exponent in front as a multiplier.

  1. Negative Argument Rule: loga(1x)=loga(x)\log_a\left(\frac{1}{x}\right) = -\log_a(x)

The logarithm of the reciprocal of a number equals the negative logarithm of the number.

  1. Logarithm of 1: loga(1)=0\log_a(1) = 0

This is because a0=1a^0 = 1 for any base aa.

  1. Logarithm Identity: aloga(x)=xa^{\log_a(x)} = x

Raising the base aa to the logarithm of xx returns xx.

  1. Inverse Property: loga(ax)=x\log_a(a^x) = x

The logarithm of a base raised to a power gives the power itself.


loga(xy)=logax+logay\log_a(xy)=\log_ax+\log_ay loga(xq)=qlogax\log_a\left(x^q\right)=q\log_ax loga(1x)=logax\log_a\left(\tfrac{1}{x}\right)=-\log_ax alogax=xa^{\log_ax}=x loga(xy)=logaxlogay\log_a\left(\tfrac{x}{y}\right)=\log_ax-\log_ay loga1=0\log_a1=0 loga(ax)=x\log_a\left(a^x\right)=x

Example

infoNote

Evaluate the logb(1b2)\log_b\left(\tfrac{1}{b^2} \right).

logb(1b2)=logb(b2)loga(1x)=logax=2loga(ax)=x \begin{align*} \log_b\left(\frac{1}{b^2} \right) &= -\log_b\left(b^2 \right) & \text{\footnotesize\textcolor{gray}{\(\log_a\left(\tfrac{1}{x}\right)=-\log_ax \)}} \\\\ &= -2 & \text{\footnotesize\textcolor{gray}{\(\log_a\left(a^x\right)=x \)}} \\\\ \end{align*}

Summary:

  • Logarithms are the inverse of exponents.
  • These laws are crucial for solving logarithmic and exponential equations.
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