1. ## Gg

$
4^\times=27
$

$
9^y=8
$

x.y=?

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

$
9^y=8
$

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

$\log 4^x=\log27$

$x\log4=\log 27$

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

$x\approx2.3774$

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

$
9^y=8
$

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

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

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

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

hence $xy=\ldots$

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

$\log 4^x=\log27$

$x\log4=\log 27$

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

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

$9^y=8$

$\log 9^y=\log8$

$y\log9=\log 8$

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

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

5. Hello, OPETH!

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

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

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

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

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

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

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