How can I isolate $x$ in this equation?

Question

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?

Answer 1

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 $y=1$.

Answer 2

You have a mistake in first equation, it should be

$$x-1 = y+2$$

from which we have $x=y+3.$ For second equation we have $$3^{y+3}+3^y=84$$

Answer 3

First multiply equation $(i)$ by $7$ to get rid of the confusing $-1$ in the exponent and obtain $$7^x=7^{y+3}$$

Then for the second part, note that $$3^{y+3}+3^y=3^y\left(3^3+3^0\right)$$

These two key steps should make a solution rather easy.

Answer 4

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.