SIMULTANEOUS EXPONENTIAL EQUATION

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In my previous post, we discussed simultaneous equations, today, we will solidify our knowledge on simulteanous equations by looking at the exponential part of the simulteanous equation.

A simultaneous exponential equation is simply a simultaneous equation derived from an exponent.

For example, in the  following equation
$5^{x+y}=125$
$3^{x-y}=3$

The accompanying simulteanous equations can be derived
$x+y=3$
$x-y=1$

Where the value of $x$ and $y$ that satisfy this equation is $2$ and $1$ respectively

But, How do we get the answer?
The accompanying example illustrates the steps to solving a simulteanous exponential equation.

👉 This post is a continuation of our series on indices 👈

Example 1
Find the value of $x$ and $y$ given that $4^{x+1}=64^{y-1}$, $(\frac{1}{4})^{2x}=(\frac{1}{16})^{y+2}$, 

Solution 
First, you rewrite all the number in base 4
$4^{x+1}=4^{3(y-1)}$
$4^{x+1}=4^{3y-3}$
The base is the same, hence, we equate the power
$x+1=3y-3$
$x-3y=-3-1$
$x-3y=-4$........eqn 1

$4^{-1(2x)}=4^{-2(y+2)}$
$4^{-2x)}=4^{-2y+4}$

The base is the same, we equate the power.
$-2x=-2y+4$
$-2x+2y=4$........eqn 2

Let's make x the subject of the formula in Eqn 1.

$x=-4+3y$

Substituting the value of $x$ in Eqn 2
$-2(-4+3y)+2y=-4$
$8-6y+2y=-4$
$8-4y=-4$
$8+4=4y$
$12=4y$
$\frac{12}{4}=\frac{4y}{4}$
$3=y$
$y=3$

Substituting y in eqn $x=-4+3y$
$x=-4+3(3)$
$x=-4+9$
$x=5$

In the above example, we solve this simultaneous using the substitution method, if you're having not yet familiar with the simultaneous equation by substitution, see this post

Example 2
If $9^{x-1}=81^{1-y}$, $25^x=5^{3-y}$, find $x$ and $y$

Solution
Let's rewrite each equation so that both sides of each equation has the same base.

$9^{x-1}=9^{2(1-y)}$
$9^{x-1}=9^{2-2y)}$

The base is the same, we equate the power.

$x-1=2-2y$
$x+2y=2+1$
$x+2y=3$......eqn 1

Let's do the same for the other eqn
$25^x=5^{3-y}$
$5^{2(x)}=5^{3-y}$
$5^{2x}=5^{3-y}$

Equating the power
$2x=3-y$
$2x+y=3$.....eqn 2

Making $x$ the subject of the formula In equation 1

$x=3-2y$

Substituting $x$ in eqn 2
$2(3-2y)+y=3$
$6-4y+y=3$
$6-3y=3$
$6-3=3y$
$3=3y$

Divide both side by $3$
$\frac{3}{3}=\frac{3y}{3}$
$1=y$
$y=1$

Substituting the value if $y$ in $x=3-2y$
$x=3-2(1)$
$x=1$

Example 3
Solve for $x$ and $y$ in the equation $3^{x-y}=27$, $3^{3-y}=2187$

Solution
$3^{x-y}=3^3$
$3^{3-y}=3^7$

Equating the powder
$x-y=3$...... eqn 1

$-y=7-3$
$-y=4$
$\frac{-y}{-1}=\frac{4}{-1}$
$y=-4$

Substituting the value of y in Eqn 1

$x-(-4)=3$
$x+4=3$
$x=3-4$
$x=-1$

Example 4
Solve $9^{1-x}=27^y$, $x-y=\frac{-3}{2}$

Solution
Expressing $9^{1-x}=27^y$ in the base of 3
$3^{2(1-x)}=3^{3(y)}$
$3^{2-2x}=3^{3y}$.

As usual, we equate the exponent
$2-2x=3y$
$-2x-3y=-2$......eqn 1

$x-y=\frac{-3}{2}$....eqn 2

Let make $x$ the subject of the formula in eqn 2
$x=\frac{-3}{2}+y$
$x=\frac{-3+2y}{2}$

Substituting the value of $x$ in eqn 1
$-2(\frac{-3+2y}{2})-3y=-2$
$\require{cancel}\bcancel{-2}((\frac{-3+2y}{\require{cancel}\bcancel{2}}-3y=-2$
$-1(-3+2y)-3y=-2$
$3-2y-3y=-2$
$-5y=-2-3$
$-5y=-5$

Divide both side by $-5$
$\frac{-5y}{-5}=\frac{-5}{-5}$
$y=1$

Substituting $y$ in $x=\frac{-3+2y}{2}$
$x=\frac{-3+2(1)}{2}$
$x=\frac{-3+2}{2}$
$x=\frac{-1}{2}$

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Example 5
Solve the simulteanous exponential equation $3^x-2^{y+2}=49$, $2^y-3^{x-2}=-1$

Solution:
Unlike the previous one, here, the base is not the same, so, let's try to make the two equations look similar.

$3^x-2^{y+2}=49$
This can be rewritten as
$3^x-2^y(2^2)=49$
$3^x-2^y(4)=49$.....eqn 1

$2^y-3^{x-2}=-1$
This can be rewritten as
$2^y-\frac{3^x}{3^2}=-1$
$2^y-\frac{3^x}{9}=-1$.....eqn 2

Let $2^y=k$, $3^x=l$
$3^x-2^y(4)=49$
$l-k(4)=49$
$l-4k=49$......eqn 3

$2^y-\frac{3^x}{9}=-1
$k-\frac{l}{9}=-1$

Taking the L.C.M of both side
$\frac{9k-l}{9}=-1$

Cross multiply 
$-l+9k=-9$....eqn 4

We are going to be solving this equation by elimination. Add eqn 3 and 4

l-4k=49
+(-l+9k=-9)
$5k=40$

Divide both side by 5
$\frac{5k}{5}=\frac{40}{5}$
$k=8$

Substituting k in eqn 3
$l-4(8)=49$
$l-32=49$
$l=81$

Recalled that $2^y=k$, $3^x=l$
$2^y=8$
$2^y=2^3$
$y=3$

$3^x=81$
$3^x=3^4$
$x=4$

Finally, we ended our series on indices. I believe you should be able to solve any question on indices. But still, if you have got a question, do well to ask our telegram community.
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