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.

lab bhattacharjee · Accepted Answer · 2014-08-17T05:58:07.470

3

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

edited Aug 17 '14 at 05:58

answered Aug 17 '14 at 05:52

lab bhattacharjee

3

Alternatively, $(5^{x+1})^2-5(5^{x+1})+6 = 0$. – JimmyK4542 Aug 17 '14 at 05:54
@JimmyK4542, Not much difference, right? – lab bhattacharjee Aug 17 '14 at 05:58
It does make factoring a bit easier, but no it isn't significantly easier. – JimmyK4542 Aug 17 '14 at 06:00
So would the answers be -0.569 and -0.317 – Issy Aug 17 '14 at 06:14
To a certain degree of approximation, yes. – Yoni Rozenshein Aug 17 '14 at 07:08

score 0 · Answer 2 · answered Aug 17 '14 at 07:28

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.

score 0 · Answer 3 · answered Aug 17 '14 at 18:16

0

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

answered Aug 17 '14 at 18:16

Bill Dubuque

3 Answers3