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Study Guides > College Algebra

Expand and Condense Logarithms

Learning Objectives

  • Expand a logarithm using a combination of logarithm rules
  • Condense a logarithmic expression into one logarithm
Taken together, the product rule, quotient rule, and power rule are often called "laws of logs." Sometimes we apply more than one rule in order to simplify an expression. For example:

logb(6xy)=logb(6x)−logby=logb6+logbx−logby\begin{array}{l}{\mathrm{log}}_{b}\left(\frac{6x}{y}\right)\hfill & ={\mathrm{log}}_{b}\left(6x\right)-{\mathrm{log}}_{b}y\hfill \\ \hfill & ={\mathrm{log}}_{b}6+{\mathrm{log}}_{b}x-{\mathrm{log}}_{b}y\hfill \end{array}

We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:

logb(AC)=logb(AC−1)=logb(A)+logb(C−1)=logbA+(−1)logbC=logbA−logbC\begin{array}{l}{\mathrm{log}}_{b}\left(\frac{A}{C}\right)\hfill & ={\mathrm{log}}_{b}\left(A{C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(A\right)+{\mathrm{log}}_{b}\left({C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}A+\left(-1\right){\mathrm{log}}_{b}C\hfill \\ \hfill & ={\mathrm{log}}_{b}A-{\mathrm{log}}_{b}C\hfill \end{array}

We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product. With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm.

Example: Using a combination of the rules for logarithms to expand a logarithm

Rewrite ln(x4y7)\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right) as a sum or difference of logs.

Answer: First, because we have a quotient of two expressions, we can use the quotient rule: ln(x4y7)=ln(x4y)−ln(7)\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)=\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right) Then seeing the product in the first term, we use the product rule: ln(x4y)−ln(7)=ln(x4)+ln(y)−ln(7)\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)=\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right) Finally, we use the power rule on the first term: ln(x4)+ln(y)−ln(7)=4ln(x)+ln(y)−ln(7)\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)=4\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)

Try It

Expand log(x2y3z4)\mathrm{log}\left(\frac{{x}^{2}{y}^{3}}{{z}^{4}}\right).

Answer: 2logx+3logy−4logz2\mathrm{log}x+3\mathrm{log}y - 4\mathrm{log}z

Q & A

Can we expand ln(x2+y2)\mathrm{ln}\left({x}^{2}+{y}^{2}\right)?

No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.
Now we will provide some examples that will require careful attention.

Example: Expanding Complex Logarithmic Expressions

Expand log6(64x3(4x+1)(2x−1)){\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right).

Answer: We can expand by applying the Product and Quotient Rules. log6(64x3(4x+1)(2x−1))=log664+log6x3+log6(4x+1)−log6(2x−1)Apply the Quotient Rule.=log626+log6x3+log6(4x+1)−log6(2x−1)Simplify by writing 64 as 26.=6log62+3log6x+log6(4x+1)−log6(2x−1)Apply the Power Rule.\begin{array}{l}{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)\hfill & ={\mathrm{log}}_{6}64+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Quotient Rule}.\hfill \\ \hfill & ={\mathrm{log}}_{6}{2}^{6}+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & {\text{Simplify by writing 64 as 2}}^{6}.\hfill \\ \hfill & =6{\mathrm{log}}_{6}2+3{\mathrm{log}}_{6}x+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Power Rule}.\hfill \end{array}

Try It 8

Expand ln((x−1)(2x+1)2(x2−9))\mathrm{ln}\left(\frac{\sqrt{\left(x - 1\right){\left(2x+1\right)}^{2}}}{\left({x}^{2}-9\right)}\right).

Answer: 12ln(x−1)+ln(2x+1)−ln(x+3)−ln(x−3)\frac{1}{2}\mathrm{ln}\left(x - 1\right)+\mathrm{ln}\left(2x+1\right)-\mathrm{ln}\left(x+3\right)-\mathrm{ln}\left(x - 3\right)

Example: Condensing Complex Logarithmic Expressions

Condense log2(x2)+12log2(x−1)−3log2((x+3)2){\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x - 1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right).

Answer: We apply the power rule first: log2(x2)+12log2(x−1)−3log2((x+3)2)=log2(x2)+log2(x−1)−log2((x+3)6){\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x - 1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right)={\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right) Next we apply the product rule to the sum: log2(x2)+log2(x−1)−log2((x+3)6)=log2(x2x−1)−log2((x+3)6){\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right) Finally, we apply the quotient rule to the difference: log2(x2x−1)−log2((x+3)6)=log2x2x−1(x+3)6{\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\frac{{x}^{2}\sqrt{x - 1}}{{\left(x+3\right)}^{6}}

Example: Rewriting as a Single Logarithm

Rewrite 2logx−4log(x+5)+1xlog(3x+5)2\mathrm{log}x - 4\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right) as a single logarithm.

Answer: We apply the power rule first: 2logx−4log(x+5)+1xlog(3x+5)=log(x2)−log((x+5)4)+log((3x+5)x−1)2\mathrm{log}x - 4\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right)=\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right) Next we apply the product rule to the sum: log(x2)−log((x+5)4)+log((3x+5)x−1)=log(x2)−log((x+5)4(3x+5)x−1)\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)=\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}{\left(3x+5\right)}^{{x}^{-1}}\right) Finally, we apply the quotient rule to the difference: log(x2)−log((x+5)4(3x+5)x−1)=log(x2(x+5)4((3x+5)x−1))\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}{\left(3x+5\right)}^{{x}^{-1}}\right)=\mathrm{log}\left(\frac{{x}^{2}}{{\left(x+5\right)}^{4}\left({\left(3x+5\right)}^{{x}^{-1}}\right)}\right)

Try It

Rewrite log(5)+0.5log(x)−log(7x−1)+3log(x−1)\mathrm{log}\left(5\right)+0.5\mathrm{log}\left(x\right)-\mathrm{log}\left(7x - 1\right)+3\mathrm{log}\left(x - 1\right) as a single logarithm.

Answer: log(5(x−1)3x(7x−1))\mathrm{log}\left(\frac{5{\left(x - 1\right)}^{3}\sqrt{x}}{\left(7x - 1\right)}\right)

Condense 4(3log(x)+log(x+5)−log(2x+3))4\left(3\mathrm{log}\left(x\right)+\mathrm{log}\left(x+5\right)-\mathrm{log}\left(2x+3\right)\right).

Answer: logx12(x+5)4(2x+3)4\mathrm{log}\frac{{x}^{12}{\left(x+5\right)}^{4}}{{\left(2x+3\right)}^{4}}; this answer could also be written log(x3(x+5)(2x+3))4\mathrm{log}{\left(\frac{{x}^{3}\left(x+5\right)}{\left(2x+3\right)}\right)}^{4}.

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