Solve exponential equation

[Virgin Vanilla]

Can someone please check my answer to the below question?

[MATH]3^{2x}-4.(3^{x})+3=0 [/MATH]
Since the base is same;

[MATH] 2x+x+1=1\\ 3x=1-1\\ x=0/3\\ x=0[/MATH]
I took base as 3 regardless of 4. Is it correct?

 

[Deleted member 4993]

Can someone please check my answer to the below question?

[MATH]3^{2x}-4.(3^{x})+3=0 [/MATH]
Since the base is same;

[MATH] 2x+x+1=1\\ 3x=1-1\\ x=0/3\\ x=0[/MATH]
I took base as 3 regardless of 4. Is it correct?

[MATH]3^{2x}-4.(3^{x})+3=0 [/MATH]
[MATH](3^{x})^2-4.(3^{x})+3=0 [/MATH]
Substitute:

u = 3^x

we have now a Quadratic Equation

u² - 4*u + 3 = 0

continue....

 

[Virgin Vanilla]

[MATH]u2 - 4*u + 3 = 0[/MATH][MATH]u-3=0[/MATH] or [MATH]u-1=0[/MATH][MATH]u=3[/MATH] or [MATH]u=1\\\\[/MATH]if [MATH] u=3[/MATH][MATH]3^x=3[/MATH][MATH]x=1\\\\[/MATH]if [MATH]u=1[/MATH][MATH]3^x=1\\ x=1[/MATH]

 

[Harry_the_cat]

Think about your last line again.

 

[HallsofIvy]

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Can someone please check my answer to the below question?

[MATH]3^{2x}-4.(3^{x})+3=0 [/MATH]
Since the base is same;

[MATH] 2x+x+1=1[/math]

No, this step is not valid. Instead, let [math]y= 3^x[/math] so the equation becomes
[math]y^2- 4y+ 3= (y- 3)(y- 1)= 0[/math].

\(\displaystyle y= 3^x= 3\) which means x= 1. This is what you had before.
\(\displaystyle y= 3^x= 1\). Before, and this is what Harry-the Cat called to your attention, you had x= 1 which is not correct. Any number to the 0 power is 1 so \(\displaystyle 3^x= 1\) leads to x= 0.

This is simply w
[math] 3x=1-1\\ x=0/3\\ x=0[/MATH]
I took base as 3 regardless of 4. Is it correct?

I'm not sure why you say that. Every x is an exponent of 3 so of course the base is 3. In any case, this equation has two roots, x= 0 and x= 1.