Proving results with logarithms

Some ques­tions give a rela­tion between let­ters and ask you to prove a log­a­rith­mic state­ment, or give a log­a­rith­mic state­ment and ask for a rela­tion between the let­ters. Both kinds use the same idea: use the laws to bring each side to a sin­gle log­a­rithm, then remove the log­a­rithms with Law 12, $\log m = \log n \implies m = n$. What is left is ordi­nary alge­bra, often an iden­tity such as $(a - b)^2 = a^2 - 2ab + b^2$.

In this les­son $\log$ with­out a base means the same base through­out, and all the let­ters stand for pos­i­tive num­bers, so that every log­a­rithm is defined.

Two ways to show that some­thing is true#

When asked to show that a state­ment holds, start with one side (usu­ally the more com­pli­cated one, often the left-hand side, LHS) and change it step by step, using the given rela­tion, until it becomes the other side (RHS). Write "= RHS" at the end.

When asked to find a rela­tion, start from the given equa­tion, com­bine each side into one log­a­rithm, drop the log­a­rithms, and sim­plify.

Exam­ple 1#

If $x^2 + y^2 = 23xy$, show that $2\log(x + y) = \log x + \log y + 2\log 5$.

Start with the left-hand side and use the power law to move the 2 inside:

$$\begin{aligned} \text{LHS} &= 2\log(x + y) \\ &= \log(x + y)^2 \\ &= \log(x^2 + y^2 + 2xy) \\ &= \log(23xy + 2xy) \\ &= \log(25xy) \\ &= \log 25 + \log x + \log y \\ &= \log 5^2 + \log x + \log y \\ &= 2\log 5 + \log x + \log y = \text{RHS} \end{aligned}$$

The given rela­tion was used in the fourth line, to replace $x^2 + y^2$ by $23xy$.

Exam­ple 2#

If $a^3 + b^3 = c^3$, show that $\log_a(c - b) + \log_a(c^2 + cb + b^2) = 3$.

From the given rela­tion, $a^3 = c^3 - b^3$. Now

$$\begin{aligned} \text{LHS} &= \log_a(c - b) + \log_a(c^2 + cb + b^2) \\ &= \log_a\left[(c - b)(c^2 + cb + b^2)\right] \\ &= \log_a(c^3 - b^3) \\ &= \log_a a^3 \\ &= 3\log_a a = 3 = \text{RHS} \end{aligned}$$

The third line uses the iden­tity $(c - b)(c^2 + cb + b^2) = c^3 - b^3$.

Exam­ple 3#

If $\displaystyle \log \frac{a - b}{2} = \frac{1}{2}(\log a + \log b)$, find the value of $\displaystyle \frac{a}{b} + \frac{b}{a}$.

Com­bine the right-hand side into one log­a­rithm:

$$\displaystyle \frac{1}{2}(\log a + \log b) = \frac{1}{2}\log ab = \log (ab)^{1/2} = \log \sqrt{ab}$$

So $\displaystyle \log \frac{a - b}{2} = \log \sqrt{ab}$, and by Law 12,

$$\displaystyle \frac{a - b}{2} = \sqrt{ab} \implies a - b = 2\sqrt{ab}$$

Square both sides:

$$\begin{aligned} (a - b)^2 &= 4ab \\ a^2 - 2ab + b^2 &= 4ab \\ a^2 + b^2 &= 6ab \end{aligned}$$

Divide both sides by $ab$: $\displaystyle \frac{a^2}{ab} + \frac{b^2}{ab} = 6$, that is $\displaystyle \frac{a}{b} + \frac{b}{a} = 6$.

Exam­ple 4#

If $\log(2a - 3b) = \log a - \log b$, find $a$ in terms of $b$.

The right-hand side is $\displaystyle \log \frac{a}{b}$, so $\displaystyle 2a - 3b = \frac{a}{b}$. Col­lect the terms in $a$ on one side:

$$\displaystyle \begin{aligned} 2a - \frac{a}{b} &= 3b \\ a\left(2 - \frac{1}{b}\right) &= 3b \\ a \cdot \frac{2b - 1}{b} &= 3b \\ a &= \frac{3b^2}{2b - 1} \end{aligned}$$

Exam­ple 5#

If $2\log x - 3\log y = \log k$, find $k$.

Move the num­bers inside with the power law, then use the quo­tient law:

$$\displaystyle \log x^2 - \log y^3 = \log k \implies \log \frac{x^2}{y^3} = \log k$$

So $\displaystyle k = \frac{x^2}{y^3}$.

Exam­ple 6#

If $\log_a b = c$, find $\log_{1/b} a$ in terms of $c$.

Write $\displaystyle \frac{1}{b} = b^{-1}$. By Law 5 and then Law 6,

$$\displaystyle \log_{b^{-1}} a = \frac{1}{-1}\log_b a = -\log_b a = -\frac{1}{\log_a b} = -\frac{1}{c}$$

Exam­ple 7#

If $a^2 + b^2 = 14ab$, show that $\displaystyle \log \frac{a + b}{4} = \frac{1}{2}(\log a + \log b)$.

This time start from the given rela­tion and build up the state­ment. Add $2ab$ to both sides:

$$a^2 + 2ab + b^2 = 16ab \implies (a + b)^2 = 16ab$$

Divide by 16 and take the square root (every­thing is pos­i­tive):

$$\displaystyle \left(\frac{a + b}{4}\right)^2 = ab \implies \frac{a + b}{4} = \sqrt{ab}$$

Take log­a­rithms of both sides:

$$\displaystyle \log \frac{a + b}{4} = \log (ab)^{1/2} = \frac{1}{2}\log ab = \frac{1}{2}(\log a + \log b)$$

Exam­ple 8#

If $\log_2 3 = p$, express $\log_2 72$ in terms of $p$.

Break 72 into primes: $72 = 8 \times 9 = 2^3 \times 3^2$. Then

$$\log_2 72 = \log_2 2^3 + \log_2 3^2 = 3 + 2\log_2 3 = 3 + 2p$$

Prac­tice#

1. If $3\log x + 2\log y = \log k$, find $k$.
2. If $\log x - \log y + \log z = \log k$, find $k$.
3. If $x^2 + y^2 = 47xy$, show that $2\log(x + y) = 2\log 7 + \log x + \log y$.
4. If $a^2 + b^2 = 7ab$, show that $\displaystyle \log \frac{a + b}{3} = \frac{1}{2}(\log a + \log b)$.
5. If $a^3 - b^3 = c^3$, show that $\log_c(a - b) + \log_c(a^2 + ab + b^2) = 3$.
6. If $\log(a - 2b) = \log a - \log b$, find $a$ in terms of $b$.
7. If $\log_3 2 = q$, express $\log_3 108$ in terms of $q$.
8. If $\log_x y = 4$, find $\log_{1/y} x$.
9. If $\displaystyle \log \frac{a + b}{3} = \frac{1}{2}(\log a + \log b)$, find $\displaystyle \frac{a}{b} + \frac{b}{a}$.

Answers#