1. ## Gg

$\displaystyle 4^\times=27$

$\displaystyle 9^y=8$

x.y=?

2. Originally Posted by OPETH
$\displaystyle 4^\times=27$

$\displaystyle 9^y=8$

x.y=?
$\displaystyle 4^x=27$

$\displaystyle \log 4^x=\log27$

$\displaystyle x\log4=\log 27$

$\displaystyle x=\frac{\log27}{\log4}$

$\displaystyle x\approx2.3774$

3. Hello,
Originally Posted by OPETH
$\displaystyle 4^\times=27$

$\displaystyle 9^y=8$

x.y=?
This can also be solved without using logarithms :

$\displaystyle \begin{cases}4^x=27 \\ 9^y=8 \end{cases} \Longleftrightarrow \begin{cases}2^{2x}=3^3 & (1) \\ 3^{2y}=2^3 & (2)\end{cases}$

From (2) we get that $\displaystyle 2=3^{\frac{2y}{3}}$ which gives, when substituting in (1) :

$\displaystyle \left(3^{\frac{2y}{3}}\right)^{2x}=3^3 \Longleftrightarrow 3^{\frac{4xy}{3}}=3^3$

hence $\displaystyle xy=\ldots$

4. Originally Posted by masters
$\displaystyle 4^x=27$

$\displaystyle \log 4^x=\log27$

$\displaystyle x\log4=\log 27$

$\displaystyle x=\frac{\log27}{\log4}$

$\displaystyle x\approx2.3774$
I obviously misread the exercise, but I'll finish up the mess that I started to hopefully arrive at the same conclusion as flyingsquirrel.

$\displaystyle 9^y=8$

$\displaystyle \log 9^y=\log8$

$\displaystyle y\log9=\log 8$

$\displaystyle y=\frac{\log8}{\log9}$

$\displaystyle xy=\frac{\log27}{\log4}\cdot\frac{\log8}{log9}=\bo xed{2.25}$

5. Hello, OPETH!

Here's a slight variation of flyingsquirrel's solution . . .

Given: .$\displaystyle \begin{array}{ccc}4^x &=& 27 \\ 9^y &=& 8 \end{array}$

. . . . . .$\displaystyle xy\:=\:?$

We have: .$\displaystyle 4^x\:=\:27\quad\Rightarrow\quad(2^2)^x \:=\:27 \quad\Rightarrow\quad 2^{2x} \:=\:3^3 \quad\Rightarrow\quad 2 \:=\:3^{\frac{3}{2x}}\;\;{\color{blue}[1]}$

. . . . . . . .$\displaystyle 9^y \:=\:8 \quad\Rightarrow\quad (3^2)^y \:=\:8 \quad\Rightarrow\quad 3^{2y} \:=\:2^3 \quad\Rightarrow\quad 2 \;=\:3^{\frac{2y}{3}}\;\;{\color{blue}[2]}$

Equate [1] and [2]: .$\displaystyle 3^{\frac{3}{2x}}\:=\:3^{\frac{2y}{3}} \quad\Rightarrow\quad \frac{3}{2x}\:=\:\frac{2y}{3}\quad\Rightarrow\quad 4xy \:=\:9$

Therefore: .$\displaystyle \boxed{xy \:=\:\frac{9}{4}}$