MHB Can the following be solved algebraically?

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The equation 3^(2x) = 12x - 3 has solutions at x = 1/2 and x = 1, confirmed by the intersection of the graphs of y = 3^(2x) and y = 12x - 3. To derive these solutions algebraically, one approach involves rewriting the equation as 3^x * 3^x = 3^1(4x - 1), leading to x = 1 as a solution. Another method is to express 3^x = √(12x - 3) and set 3^x equal to 3^(1/2), which yields x = 1/2. Both methods validate the original solutions through algebraic manipulation.
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I noticed the solution to

$$3^{2x} = 12x-3$$ is $$ x=\frac{1}{2},1$$

is there a way I can arrive here with a bit of simple algebra?
 
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Bushy said:
I noticed the solution to

$$3^{2x} = 12x-3$$ is $$ x=\frac{1}{2},1$$

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.
 
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