Solve $5^{2x+2}-5^{x+2}+6=0 $

Question

How do we solve $5^{2x+2}-5^{x+2}+6=0 $? I know I have to use logarithms but I am not sure how to do it.

Answer 1

Using $\displaystyle a^{mx+n}=a^n(a^x)^m,$

we have $$25(5^x)^2-25(5^x)+6=0$$ which is a Quadratic Equation in $5^x$

$$5^x=\frac{25\pm5}{50}=\frac25,\frac35$$

Taking logarithm, $$x\log 5=\log2-\log5,\log3-\log5$$

Reference : Exponent Combination Laws

Answer 2

Let

$5^{x+2}=y$

$\implies 25\times5^x=y$

$\implies 5^x=\frac{y}{25}$

$\implies 5^{2x}=\frac{y^2}{625}$

$\implies 5^{2x+2}=\frac{y^2}{25}$

The equation then, shall reduce to

$\frac{y^2}{25}-y+6=0$

$\implies y^2-25y+150=0$

This can be solved using quadratic equations.

Answer 3

Hint $\ $ For $\,X = 5^{x+1}\,$ the equation is $\, 0\, =\, X^2 - 5 X + 6\, =\, (X-2)(X-3)$