EXPONENTIAL QUADRATIC EQUATIONS

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An exponential quadratic equation is an equation where the variable we are solving appears in the exponent so that we solve a quadratic equation.

👉This post extends our series on indices 👈

If an exponential equation has the same base on each side of the equation, then the exponent is equal. 

Using this fact, we can solve the following

Example 1
Solve for $x$ in the equation
$7^{5x+8}=7^{x^2+3x}$

Solution: 
The base($7$) is the same on both sides, hence the exponents are equal. 
$5x+8=x^2+3x$

Re-arranging the equation
$0=x^2+3x-5x-8$
$0=x^2-2x-8$
$x^2-2x-8=0$

$x^2-4x+2x-8=0$
$x(x-4)+2(x-4)=0$
$(x+2)(x-4)=0$
$x=-2$ or $x=4$

Example 2
Sole for $x$ in the equation
$4^x-2^{x+2}=32$

Solution:
Simplifying  the indicial equation
$2^{2(x)}-2^x\times2^2=32$
$2^{x(2)}-2^x(4)=32$
Let $2^x=t$
$t^2-t(4)=32$
$t^2-4t-32=0$

By factorization
$t^2+4t-8t-32=0$
$t(t+4)-8(t+4)=0$
$(t-8)(t+4)=0$
$t=8$ or $-4$

$t$ can't be negative, hence
$t=8$, but $t≠-4$

Inserting the value of $t$ in $2^x=t$
$2^x=8$
$2^x=2^3$

The base is the same, therefore,
$x=3$

Example 3
Find the value of x in this equation
$5^{2x-3}-126(5^{x-3})+1=0$

Solution
$5^{2x-3}-126(5^{x-3})+1=0$

will translate to
$\frac{5^{2x}}{5^3}-\frac{126(5^x)}{5^3}+1=0$
$\frac{5^{x(2)}}{125}-\frac{126(5^x)}{125}+1=0$

Let $5^x=t$
$\frac{t^2}{125}-\frac{126t}{125}+1=0$

Taking the L.C.M
$\frac{t^2-126t+125}{125}=0$

Cross multiply
$t^2-126t+125=0\times125$
$t^2-126t+125=0$

Solving quadratically
$t^2-125t-1t+125=0$
$t(t-125)-1(t-125)=0$
$(t-1)(t-125)=0$
$t=1$ or $125$

Substituting the value of t in $5^x=t$
When t=1
$5^x=1$
$5^x=5^0$
$x=0$

When t=125
$5^x=125$
$5^x=5^3$
$x=3$

Therefore $x=3$ and $0$

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Example 4
Solve for x if $2^{2x+1}+2^x-1=0$

Solution:
$2^{2x}\times2+2^x-1=0$
$2^{x(2)}\times2+2^x-1=0$

Let $2^x=t$
$t^2(2)+t-1=0$
$2t^2+t-1=0$

By factorization
$2t^2+2t-t-1=0$
$2t(t+1)-1(t+1)=0$
$(2t-1)(t+1)=0$
$2t-1=0$ or $t+1=0$
$2t=1$ or $t=-1$
$\frac{2t}{2}=\frac{1}{2}$ or $t=-1$
$t=\frac{1}{2}$ or $t=-1$

$t$ cannot be negative, therefore,
$t=\frac{1}{2}$ but $t≠-1$

Inserting the value of t in $2^x=t$
$2^x=\frac{1}{2}$
$2^x=2^{-1}$
$x=-1$


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