<s>[INST] How can I isolate $x$ in this equation? Simultaneous equations - $7^{x-1} = 49 \hspace{1 mm} (7^{y})$ --------- (i)
$\hspace{46 mm} 3^x+3^y = 84$ ---------- (ii)
The first I make x subject of equation in this way:
$\Longrightarrow 7^{x-1} = 49 \hspace{1 mm} (7^{y})$
$\Longrightarrow 7 \times 7^x = 7^2 \times 7^y$
$\Longrightarrow 7^x = 7^{y+1}$
$\Longrightarrow x = y + 1$
The second I tried with logarithms but could not separate x in simple form:
$\Longrightarrow 3^x + 3^y = 84$
$\Longrightarrow 3^x = 84 - 3^y$
$\Longrightarrow \lg3^x = \lg (84 - 3^y)$
$\Longrightarrow x \lg3 = \lg(84 - 3^y)$
$\Longrightarrow x = \dfrac{\lg(84 - 3^y)}{\lg3}$
How can I simplify this further?
 [/INST] You have made an error in the first equation:
$\Rightarrow 7^{x-1}=49\hspace{1 mm}(7^{y})$
$\Rightarrow \dfrac{7^{x}}{7}=49\hspace{1 mm}(7^{y})$
$\Rightarrow 7^{x}=7^{3}\hspace{1 mm}(7^{y})$
$\Rightarrow 7^{x}=7^{y+3}$
$\Rightarrow x=y+3$
Then, we can substitute this expression for $x$ into the second equation:
$\Rightarrow 3^{x}+3^{y}=84$
$\Rightarrow 3^{y+3}+3^{y}=84$
$\Rightarrow 27\cdot{3^{y}}+3^{y}=84$
$\Rightarrow 28\cdot{3^{y}}=84$
$\Rightarrow {3^{y}}=3^{1}$
$\Rightarrow y=1$
Now, let's substitute this value of $y$ into the first equation:
$\Rightarrow x=(1)+3$
$\hspace{8.75 mm}=4$
Therefore, $x=4$ and $y=1$.
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