<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] Your solution is incorrect.
It should be as follows:
$$7^{x-1}=49\cdot 7^y$$
$$ \implies 7^{x-1}=7^{y+2}$$
Comparing both sides of the equation, we get
$$x-1=y+2$$
$$ \implies x=y+3\tag1$$
From the second equation, we have
$$3^x+3^y=84$$
$$ \implies 3^{y+3}+3^y=84$$
$$ \implies 3^y(3^3+1)=84$$
$$ \implies 3^y\cdot 28=84$$
$$ \implies 3^y=3=3^1$$
Again comparing both sides of the equation, we get
$$y=1$$
And using $(1)$, we get
$$x=4$$
Hence the required solution is
$$\boxed{x=4,y=1}$$
Hope this helps you.
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