Riya Hansen

2022-07-02

Need help with solving logarithmic equations ${2}^{2x}-{2}^{x}-6=0$ and ${3}^{2x}-5\cdot {3}^{x}+4=0$
No clue how to approach this problem..

Caiden Barrett

Expert

Hint:
In each case, you have quadratics in ${b}^{x}$:
In the first, put $\phantom{\rule{thickmathspace}{0ex}}y={2}^{x}$. Then you have ${y}^{2}-y-6=0$
For the second, put $\phantom{\rule{thickmathspace}{0ex}}y={3}^{x}$. Then you have ${y}^{2}-5y+4=0$
Note: Both quadratic equations factor quite nicely.

We can see immediately that in both solutions, ${2}^{x}$ must necessarily be positive, since $2>0$. So we're left with
$\begin{array}{}\text{(1)}& {2}^{x}=3\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}\left({2}^{x}\right)=\mathrm{ln}3\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x\mathrm{ln}2=\mathrm{ln}3\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x=\frac{\mathrm{ln}3}{\mathrm{ln}2}\end{array}$

So we have
$\begin{array}{}\text{(1)}& {3}^{x}=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}\left({3}^{x}\right)=\mathrm{ln}1\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x\mathrm{ln}3=\stackrel{=\phantom{\rule{thinmathspace}{0ex}}0}{\stackrel{⏞}{\mathrm{ln}1}}=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x=0\end{array}$
And we have
$\begin{array}{}\text{(2)}& {3}^{x}=4\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}\left({3}^{x}\right)=\stackrel{=\mathrm{ln}\left({2}^{2}\right)=2\mathrm{ln}2}{\stackrel{⏞}{\mathrm{ln}4}}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x\mathrm{ln}3=2\mathrm{ln}2\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x=\frac{2\mathrm{ln}2}{\mathrm{ln}3}\end{array}$

Joshua Foley

Expert

a. Let ${2}^{x}=m\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{m}^{2}-m-6=0$
Solve the quadratic equation to get:

But remember ${2}^{x}=m$ so:

$\mathrm{ln}$ both sides of the equation to get:
$\mathrm{ln}{2}^{x}=\mathrm{ln}3\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x\mathrm{ln}2=\mathrm{ln}3\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=\frac{\mathrm{ln}3}{\mathrm{ln}2}$
$\mathrm{ln}{2}^{x}=\mathrm{ln}-2$
You cannot take the $\mathrm{ln}$ of a negative number so that solution is not applicable.
$\therefore x=\frac{\mathrm{ln}3}{\mathrm{ln}2}$
b. Let ${3}^{x}=n\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{n}^{2}-5n+4=0$
Solve the quadratic equation to get:

But remember ${3}^{x}=n$ so:

$\mathrm{ln}$ both sides of the equation to get:
$\mathrm{ln}{3}^{x}=\mathrm{ln}1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x\mathrm{ln}3=ln1$
but
$\mathrm{ln}1=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=0$
$\mathrm{ln}{3}^{x}=\mathrm{ln}4\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x\mathrm{ln}3=\mathrm{ln}4\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=\frac{\mathrm{ln}4}{\mathrm{ln}3}$

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