Solve the exponential equation \(\displaystyle{5}^{{{2}{x}+{1}}}+{5}^{{x}}-{4}={0}.\)

monkeyman130yb

monkeyman130yb

Answered question

2022-03-22

Solve the exponential equation
52x+1+5x4=0.

Answer & Explanation

Laylah Hebert

Laylah Hebert

Beginner2022-03-23Added 15 answers

The given equation 52x+1+5x4=0 can be rewritten as 52x5+5x4=0

This can be rewritten as 5(5x)2+5x4=0
Substituting u=5x, we get: 5u2+u4=0
Solving for u:
u=1±145(4)25
u=1±910u=45 or u=1
So,
5x=45 or 5x=1
But 5x=1 has no solution, so this case is ruled out.
We have only 5x=45. To solve for x, we take the logarithm (say the natural logarithm) on both sides, we get:
ln(5x)=ln(45)
xln(5)=ln(4)ln(5)x=ln(4)ln(5)ln(5)
The properties used here are:
ln(ab)=bln(a), and ln(ab)=ln(a)ln(b).
Hope this helps.

Aidyn Wall

Aidyn Wall

Beginner2022-03-24Added 10 answers

Hint: Let t=5x and note that
5t2+t4=(5t4)(t+1).

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