Solve the exponential equation $5^{2x+1}+5^x-4=0$.

Question

Solve the equation $$5^{2x+1}+5^x-4=0$$

I need to solve this equation using logarithms. I am in Year 10 and I am clueless on what to do.

Answer 1

Hint: Let $t=5^x$ and note that $$5t^2+t-4=(5t-4)(t+1).$$

Answer 2

Take $y = 5^x$, then the equation becomes

$$5y^2 + y - 4 = 0$$ This is a quadratic equation in $y$, now solve for feasible values of $y$ (you can't have a negative $y$ for example).

Answer 3

Use your knowledge of Laws of Exponents to rewrite the expression $$5^{2x+1}+5^x-4=0\Rightarrow 5\cdot(5^x)^2+5^x-4=0$$

Then, solve using your knowledge of quadratic equations.

$$ \begin{align*} 0 & = 5\cdot(5^x)^2+5^x-4=0 \\ 0 & = (5\cdot 5^x-4)(5^x+1) \\ \end{align*}$$

$$ \begin{align*} 5\cdot 5^x & =4 \\ 5^{x+1} & = 4 \\ x+1 &= \log_54 \\ x & = \log_54-1 \end{align*}$$ or $$5^x=-1$$ for which there is no real solution.

Answer 4

The given equation $5^{2x+1}+5^x-4=0$ can be rewritten as $5^{2x}\cdot5+5^x-4=0$

This can be rewritten as $5\cdot(5^x)^2+5^x-4=0$

Substituting $u=5^x$, we get: $5u^2+u-4=0$

Solving for $u$:

$$u=\frac{-1\pm\sqrt{1-4\cdot5\cdot(-4)}}{2\cdot5}$$

$$u=\frac{-1\pm9}{10}\implies u=\frac{4}{5} \text{ or }u=-1$$

So,

$$5^x=\frac{4}{5} \text{ or } 5^x=-1$$

But $5^x=-1$ has no solution, so this case is ruled out.

We have only $5^x=\frac{4}{5}$. To solve for $x$, we take the logarithm (say the natural logarithm) on both sides, we get:

$$\ln(5^x)=\ln(\frac{4}{5})$$ $$\implies x\ln(5)=\ln(4)-\ln(5) \implies x=\frac{\ln(4)-\ln(5)}{\ln(5)}$$

The properties used here are:

$\ln(a^b)=b\ln(a)$, and $\ln(\frac{a}{b})=\ln(a)-\ln(b)$.

Hope this helps.