Solving an exponential with three different bases

Question

$$2^x+4^x=8^x$$ Solve for $x$. I reduced the bases to $2$ and try to use logarithms but could not get passed the logarithm of an expression. When typing into online calculators they say there is no solution but graphing shows an answer.

Answer 1

Hint: $4 = 2^2$ and $8 = 2^3$.

Let $z = 2^x$. Then, using the hint, we can rewrite the original equation as \begin{align} 2^x + 4^x = 8^x &\implies 2^x + (2^2)^x = (2^3)^x \\ &\implies 2^x + (2^{x})^2 = (2^{x})^3 \\ &\implies z + z^2 = z^3. \end{align} Bringing all of the terms to one side of the equation, we obtain $$ 0 = z^3 - z^2 - z = z(z^2 - z -1). $$ One possible solution is $z = 0$, but this doesn't work, as $z = 2^x = 0$ has no real solutions. Dealing with the factor $z^2 - z - 1$, e.g. via the quadratic formula, we get $$ z = \frac{1 \pm \sqrt{1 - (4)(1)(-1)}}{2(1)} = \frac{1}{2} \pm \frac{\sqrt{5}}{2}. $$ Again, since $z = 2^x$, so $x = \log_2(z)$, and so $$ x = \log_2\left( \frac{1}{2} \pm \frac{\sqrt{5}}{2} \right). $$ However, $\log$ is undefined for negative numbers, so this reduces to only one real solution, namely $$ x = \log_2\left( \frac{1}{2} + \frac{\sqrt{5}}{2} \right). $$

Addendum: This is probably overkill for what is likely a precalculus problem, but I $\heartsuit$ the complex logarithm, so: just for funsies, note that we can deal with the logarithm of a negative number if we are willing to deal with complex numbers and some subtle issues in complex analysis. Let's suppose that $2^x = z$. Then, if we assume that $x = a+ib \in \mathbb{C}$ and that $z\in\mathbb{R}$, we have $$2^x = \mathrm{e}^{\log(2) x} = \mathrm{e^{\log(2)a + i\log(2)b}} = \mathrm{e}^{\log|z| + i2k\pi}, $$ where $\log$ the real natural logarithm (or the principle branch of the complex logarithm, if you prefer), and $k\in\mathbb{Z}$. Equating real and imaginary parts, we get $$ \begin{cases} a = \frac{\log|z|}{\log(2)} \\ b = \frac{2\pi}{\log(2)}. \end{cases} $$ Therefore any number of the form $$ \frac{\log|z|}{\log(2)} + i \frac{2k\pi}{\log(2)} $$ is a solution to the equation $2^x = z$, where $z$ is any nonzero real number (positive or negative).

In particular, we determined above that there are solutions corresponding to $$ z = \frac{1}{2} \pm \frac{\sqrt{5}}{2}. $$ Let's label these two solutions $$ \varphi = \frac{1}{2} + \frac{\sqrt{5}}{2} \qquad \text{and} \qquad \psi = \frac{1}{2} - \frac{\sqrt{5}}{2} = -\varphi^{-1}. $$ As $|\varphi| = \varphi$ and $|\psi| = \varphi^{-1}$, we may use the arguments above to conclude that the solutions to the original equation must then be elements of the set $$ \left\{ \frac{\log(\varphi)}{\log(2)} + i \frac{2k\pi}{\log(2)}, \frac{\log(\varphi^{-1})}{\log(2)} + i \frac{2k\pi}{\log(2)} \ \middle|\ k \in \mathbb{Z} \right\} = \left\{ \pm\frac{\log(\varphi)}{\log(2)} + i \frac{2k\pi}{\log(2)} \ \middle|\ k \in \mathbb{Z} \right\}. $$ There is a rather lovely kind of symmetry here, n'est-ce pas?

Answer 2

Set $y = 2^x$, then you get \begin{align} y+y^2= y^3 \ \ \implies \ \ \ y(1+y-y^2) = 0 \ \ \implies \ \ y=0 \ \ \text{ or } \ \ y = \frac{1\pm\sqrt{5}}{2}. \end{align} Hence it follows that \begin{align} y = 2^x=\frac{1+\sqrt{5}}{2} \ \ \implies \ \ x = \log_2\left(\frac{1+\sqrt{5}}{2}\right). \end{align} Note that $y \neq 0$ nor does it equal $(1-\sqrt{5})/2$.

Answer 3

Let $2^x=t$.

Thus, $$1+t=t^2,$$ or $$t=\frac{1+\sqrt{5}}{2},$$ which gives $$x=\log_2\frac{1+\sqrt{5}}{2}.$$

Answer 4

Let $2^x=y$.

The equation becomes $$y+y^2=y^3\\ 1+y=y^2\\ y^2-y-1=0\\ y=\frac{1\pm\sqrt{5}}{2}\\ 2^x=\frac{1\pm\sqrt{5}}{2}\\ x=\log_2{\frac{1\pm\sqrt{5}}{2}}$$

Because $\frac{1-\sqrt{5}}{2}<0$ the logarithm of a negative number does not exist,

$$x=\log_2{\frac{1+\sqrt{5}}{2}}\text.$$

Answer 5

Just to be different, divide by $\,4^x \ne 0\,$, first, then look at the resulting equation $\,\frac{1}{2^x}+1=2^x\,$ and compare it with the famous one for the golden ratio: $\,\frac{1}{\varphi}+1=\varphi\,$.