Equal powers and the k method

Some ques­tions give a chain of equal pow­ers, such as $2^{x} = 3^{y} = 12^{z}$, and ask you to prove a rela­tion between $x$, $y$ and $z$. You can­not find the three num­bers them­selves, but you do not need to. The trick is to call the com­mon value $k$ and write each base as a power of $k$. This les­son shows the trick on the stan­dard ques­tions.

The k method#

1. Put every power in the chain equal to one let­ter: $2^{x} = 3^{y} = 12^{z} = k$.
2. From each equa­tion, write the base as a power of $k$, using Law 14: if $2^{x} = k$ then $2 = k^{1/x}$.
3. Find a link between the bases: for exam­ple $12 = 2^{2} \times 3$.
4. Replace each base by its power of $k$, col­lect the pow­ers of $k$ with the laws, and com­pare expo­nents with Law 11.

A small but impor­tant step in the method is writ­ing a num­ber such as $12$, $75$ or $6$ as a prod­uct of the other bases. Break it into primes first.

Exam­ple 1: a cycle of pow­ers#

If $a^{x} = b$, $b^{y} = c$ and $c^{z} = a$, prove that $xyz = 1$.

This one needs no $k$. Start with the first equa­tion and keep replac­ing a let­ter by the power that equals it.

$$\begin{aligned} a^{x} &= b \\ (c^{z})^{x} &= b \quad \text{(since } a = c^{z}\text{)} \\ c^{zx} &= b \\ (b^{y})^{zx} &= b \quad \text{(since } c = b^{y}\text{)} \\ b^{xyz} &= b^{1} \end{aligned}$$

Com­par­ing expo­nents, $xyz = 1$.

Exam­ple 2#

If $2^{x} = 3^{y} = 12^{z}$, show that $\displaystyle \frac{1}{z} = \frac{1}{y} + \frac{2}{x}$.

Let $2^{x} = 3^{y} = 12^{z} = k$. Then

$$2 = k^{1/x}, \qquad 3 = k^{1/y}, \qquad 12 = k^{1/z}$$

Now $12 = 2 \times 2 \times 3 = 2^{2} \times 3$. Replace each num­ber by its power of $k$:

$$\displaystyle \begin{aligned} 2^{2} \times 3 &= k^{1/z} \\ (k^{1/x})^{2} \times k^{1/y} &= k^{1/z} \\ k^{2/x} \times k^{1/y} &= k^{1/z} \\ k^{\frac{2}{x} + \frac{1}{y}} &= k^{1/z} \end{aligned}$$

Com­par­ing expo­nents, $\displaystyle \frac{2}{x} + \frac{1}{y} = \frac{1}{z}$, which is what we had to show.

Exam­ple 3: a neg­a­tive expo­nent in the chain#

If $2^{x} = 3^{y} = 6^{-z}$, show that $\displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.

Let $2^{x} = 3^{y} = 6^{-z} = k$. Then

$$2 = k^{1/x}, \qquad 3 = k^{1/y}, \qquad 6 = k^{1/(-z)} = k^{-1/z}$$

Since $6 = 2 \times 3$:

$$\displaystyle \begin{aligned} k^{1/x} \times k^{1/y} &= k^{-1/z} \\ k^{\frac{1}{x} + \frac{1}{y}} &= k^{-1/z} \\ \frac{1}{x} + \frac{1}{y} &= -\frac{1}{z} \\ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} &= 0 \end{aligned}$$

Exam­ple 4: a link given in the ques­tion#

If $a^{x} = b^{y} = c^{z}$ and $b^{2} = ac$, show that $\displaystyle y = \frac{2xz}{x + z}$.

Let $a^{x} = b^{y} = c^{z} = k$. Then $a = k^{1/x}$, $b = k^{1/y}$ and $c = k^{1/z}$. The link between the bases is given: $b^{2} = ac$.

$$\displaystyle \begin{aligned} (k^{1/y})^{2} &= k^{1/x} \cdot k^{1/z} \\ k^{2/y} &= k^{\frac{1}{x} + \frac{1}{z}} \\ \frac{2}{y} &= \frac{1}{x} + \frac{1}{z} \\ \frac{2}{y} &= \frac{z + x}{xz} \end{aligned}$$

Turn both sides upside down: $\displaystyle \frac{y}{2} = \frac{xz}{z + x}$, so $\displaystyle y = \frac{2xz}{x + z}$.

Exam­ple 5#

If $3^{x} = 5^{y} = 75^{z}$, show that $\displaystyle z = \frac{xy}{2x + y}$.

Let $3^{x} = 5^{y} = 75^{z} = k$. Then $3 = k^{1/x}$, $5 = k^{1/y}$ and $75 = k^{1/z}$. Break $75$ into primes: $75 = 3 \times 25 = 3 \times 5^{2}$.

$$\displaystyle \begin{aligned} 3 \times 5^{2} &= k^{1/z} \\ k^{1/x} \cdot (k^{1/y})^{2} &= k^{1/z} \\ k^{\frac{1}{x} + \frac{2}{y}} &= k^{1/z} \\ \frac{1}{x} + \frac{2}{y} &= \frac{1}{z} \\ \frac{y + 2x}{xy} &= \frac{1}{z} \end{aligned}$$

Turn­ing both sides upside down, $\displaystyle z = \frac{xy}{2x + y}$.

Exam­ple 6: a ratio#

If $2^{x} = 4^{y} = 8^{z}$, find the ratio $x : y : z$.

Here all the bases are pow­ers of $2$, so no $k$ is needed: $2^{x} = 2^{2y} = 2^{3z}$, which gives $x = 2y = 3z$. If $x = 6$, then $y = 3$ and $z = 2$. So $x : y : z = 6 : 3 : 2$.

Prac­tice#

1. If $2^{x} = 3^{y} = 6^{z}$, show that $\displaystyle \frac{1}{z} = \frac{1}{x} + \frac{1}{y}$.
2. If $3^{x} = 5^{y} = 15^{z}$, show that $\displaystyle z = \frac{xy}{x + y}$.
3. If $2^{a} = 3^{b} = 18^{c}$, show that $\displaystyle \frac{1}{c} = \frac{1}{a} + \frac{2}{b}$.
4. If $4^{x} = 5^{y} = 20^{z}$, show that $\displaystyle z = \frac{xy}{x + y}$.
5. If $5^{x} = 2^{y} = 10^{-z}$, show that $\displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.
6. If $a^{x} = b^{y} = c^{z}$ and $abc = 1$, show that $\displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.
7. If $2^{x} = 8^{y} = 16^{z}$, find $x : y : z$.

Answers#