what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$

Question

I have tried to solve this expression but I'm stuck: $$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$ and since $3^{a} = 4^{b}$: $$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac{2a}{a} } = 4^{ b } + 3^{ a }$$ Firstly is this right? Secondly how to complete?

Answer 1

You were close. However, you simplified $\frac{2b}b$ into $b$ and $\frac{2a}a$ into $a$ in your final step, which is not right. Correct that mistake, and you should have your solution.

On a less serious note, I would do an extra step here, for clarity: $$ 3^{2a/b} + 4^{2b/a} = (3^a)^{2/b} + (4^b)^{2/a} = 4^{2a/a} + 3^{2b/b} $$ Also I prefer to not write fractions in exponents unless necessary. I think they're somewhat ugly.

Answer 2

Let $3^a=4^b=k^{ab}\implies 3=k^b,4=k^a$

$$9^{a/b}+16^{a/b}$$

$$=(3)^{2a/b}+4^{2b/a}$$

$$=(k^b)^{2a/b}+(k^a)^{2b/a}$$

$$=k^{2b}+k^{2a}$$

$$=(k^b)^2+(k^a)^2=?$$

Answer 3

Let $a = 1$. Then $4^b = 3 \Rightarrow b = \log_4 3 = \frac{\log 3}{\log 4}$.

Therefore:

$$9^{a/b} + 16^{b/a}$$ $$=9^{\log 4 / \log 3} + 16^{\log3 / \log 4}$$ $$=3^{2 \log 4 / \log 3} + 4^{2 \log 4 / \log 3}$$ $$=3^{2 \log_3 {4} } + 4^{2 \log_4 {3}}$$ $$= \left( 3^{\log_3 4} \right )^2 + \left( 4^{\log_4 3} \right)^2 $$

Answer 4

$$(9)^{ \displaystyle\frac{a}{b} } + (16)^{ \displaystyle\frac{b}{a} } = \left(3^{\displaystyle \color{magenta}{2}}\right)^{ \displaystyle\frac{\color{red}{a}}{b} } + \left(4^{\displaystyle \color{brown}{2}}\right)^{ \displaystyle\frac{\color{blue}{b}}{a} } =\left(3^{\displaystyle \color{red}a}\right)^{ \displaystyle\frac{\color{magenta}2}{b} } + \left(4^{\displaystyle \color{blue}b}\right)^{ \displaystyle\frac{\color{brown}2}{a} } $$

Now if $3^{\displaystyle a} = 4^{\displaystyle b}$, then

$$\left(3^{\displaystyle a}\right)^{ \displaystyle \frac{2}{b} } + \left(4^{\displaystyle b}\right)^{ \displaystyle \frac{2}{a} } = \left(4^{\displaystyle b}\right)^{ \displaystyle \frac{2}{b} } + \left(3^{\displaystyle a}\right)^{ \displaystyle \frac{2}{a} } = 25 $$

The rules are:

• If you have three numbers $x,y$ and $z$, then $(x^y)^z = x^{y~\cdot ~z}= x^{z~\cdot~ y}=(x^z)^y$.
• If you have three numbers $x,y$ and $z\neq 0$, then $\frac{x~\cdot ~y}{z} = x~\cdot~ \frac{y}{z}= y~\cdot~ \frac{x}{z} =\frac{y~\cdot~ x}{z}$.

Answer 5

Here a short way:

• $3^a = 4^b \Leftrightarrow 3 = 4^{\frac{b}{a}}$ and $4= 3^{\frac{a}{b}}$
• $\Rightarrow 9^{\frac{a}{b}} = \left(3^{\frac{a}{b}}\right)^2 = 4^2$ and
• $\Rightarrow 16^{\frac{b}{a}} = \left(4^{\frac{b}{a}}\right)^2 = 3^2$

Now, just sum up.

Answer 6

I think the most straightforward approach is computing $a$ in terms of $b$.

$3^a=4^b$ so $a = \log_34^b=b\log_34$

$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 9^{ \log_34}+16^{\log_43}=3^{ 2\log_34}+4^{2\log_43}=16+9=25$