Can the following be solved algebraically?

[Bushy]

I noticed the solution to

\(\displaystyle 3^{2x} = 12x-3\) is \(\displaystyle x=\frac{1}{2},1\)

is there a way I can arrive here with a bit of simple algebra?

 

[anemone]

I noticed the solution to

\(\displaystyle 3^{2x} = 12x-3\) is \(\displaystyle x=\frac{1}{2},1\)

is there a way I can arrive here with a bit of simple algebra?

Hi Bushy,

Your observation is sharp and correct. We know the graph of $y=3^{2x}$ and $y=12x-3$ will intersect at most twice, and so, by observation, $x=\frac{1}{2},\,1$ are the correct $x$-coordinate of the point of intersections.

But, if you want to convince people how you arrived at the solution, you might want to try the following:

1.

$3^{2x}= 12x-3$

$3^x \cdot 3^x= 3^1(4x-1)$

If we let $3^x=3^1$ which implies $x=1$, then we see that $3^1=4(1)-1=3$ therefore $x=1$ is a solution.

1.

$3^{x}= \sqrt{12x-3}$

$3^x = \sqrt{3}\sqrt{4x-1}=3^{\frac{1}{2}}\sqrt{4x-1}$

Similarly, if we let $3^x=3^{\frac{1}{2}}$ which suggests $x=\frac{1}{2}$, we have $1=\sqrt{4(\frac{1}{2})-1}=1$, thus, $x=\frac{1}{2}$ is the other solution.