#### Virgin Vanilla

##### New member

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\(\displaystyle 3^{2x}-4.(3^{x})+3=0 \)

Since the base is same;

\(\displaystyle 2x+x+1=1\\

3x=1-1\\

x=0/3\\

x=0\)

I took base as 3 regardless of 4. Is it correct?

- Thread starter Virgin Vanilla
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\(\displaystyle 3^{2x}-4.(3^{x})+3=0 \)

Since the base is same;

\(\displaystyle 2x+x+1=1\\

3x=1-1\\

x=0/3\\

x=0\)

I took base as 3 regardless of 4. Is it correct?

- Joined
- Jun 18, 2007

- Messages
- 24,691

\(\displaystyle 3^{2x}-4.(3^{x})+3=0 \)

\(\displaystyle 3^{2x}-4.(3^{x})+3=0 \)

Since the base is same;

\(\displaystyle 2x+x+1=1\\

3x=1-1\\

x=0/3\\

x=0\)

I took base as 3 regardless of 4. Is it correct?

\(\displaystyle (3^{x})^2-4.(3^{x})+3=0 \)

Substitute:

u = 3

we have now a Quadratic Equation

u

continue....

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\(\displaystyle u-3=0\) or \(\displaystyle u-1=0\)

\(\displaystyle u=3\) or \(\displaystyle u=1\\\\\)

if \(\displaystyle u=3\)

\(\displaystyle 3^x=3\)

\(\displaystyle x=1\\\\\)

if \(\displaystyle u=1\)

\(\displaystyle 3^x=1\\

x=1\)

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Think about your last line again.

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No, this step is not valid. Instead, let \(\displaystyle y= 3^x\) so the equation becomesCan someone please check my answer to the below question?

\(\displaystyle 3^{2x}-4.(3^{x})+3=0 \)

Since the base is same;

\(\displaystyle 2x+x+1=1\)

\(\displaystyle y^2- 4y+ 3= (y- 3)(y- 1)= 0\).

\(\displaystyle y= 3^x= 3\) 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.

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.This is simply w

\(\displaystyle 3x=1-1\\

x=0/3\\

x=0\)

I took base as 3 regardless of 4. Is it correct?