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

ðŸ‘‰This post extends our series on indices ðŸ‘ˆ

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

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