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This is a math problem I'm currently working on.

$$5^{2x} + 4(5^x) - 5 = 0$$

I've used logarithm to try solve the problem. Here's what I've done so far:

\begin{align}5^{2x} + 4(5^x) - 5 &= 0\\ 5^{2x} + 5^x &= \frac{5}{4}\\ \log_5{2x} + \log_5{x} &= \log_5\left(\frac{5}{4}\right)\\ \log_5{2x^2} &= \log_5\left(\frac{5}{4}\right)\\ 5^{2x^2} &= 5^\frac{5}{4}\\ (2x^2)\log5 &= \frac{5}{4}\log5\\ 2x^2 &= \frac{5}{4}\\ 2x^2 &= 1.25\\ x^2 &= 0.625\\ \sqrt {x^2} &= \sqrt {0.625}\\ x &= \frac{\sqrt {10}}{4}\end{align}

Substituting the value for $x$ into the equation doesn't equate it to zero. I've tried several different ways but I have still not come up with a correct answer.

lioness99a
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AugieJavax98
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3 Answers3

3

Hint:

A useful trick is to substitute $u=5^x$ to obtain a quadratic equation:

$$u^2+4u-5=0$$ Can you solve for $u$, and then solve for $x$?

projectilemotion
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3

You made a pretty big mistake because $\log a + \log b \neq \log (a+b)$. So, your third line does not follow from your second.

To actually solve the problem, here's two hints:

  • Set $y=5^x$
  • Use the fact that $a^{bc} = (a^b)^c$
5xum
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2

$\require{enclose}5^{2x}+4(5)^{x}-5=0 \enclose{updiagonalstrike}{\implies} 5^{2x}+5^{x}=\frac{5}{4}$

Instead:

Let $y=5^x\quad$ noting that $y\geq 0$, then we have:

$y^2+4y-5=0\implies (y+5)(y-1)=0$

$\implies y=-5\quad\text{reject since}\quad y\geq 0\quad \text{or}\quad y=1\implies 5^x=1\implies x=0$

mrnovice
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