Solve exponential equation

Virgin Vanilla

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Apr 6, 2021
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Can someone please check my answer to the below question?

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

Since the base is same;

\(\displaystyle 2x+x+1=1\\
3x=1-1\\
x=0/3\\
x=0\)

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

Subhotosh Khan

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Can someone please check my answer to the below question?

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

Since the base is same;

\(\displaystyle 2x+x+1=1\\
3x=1-1\\
x=0/3\\
x=0\)

I took base as 3 regardless of 4. Is it correct?
\(\displaystyle 3^{2x}-4.(3^{x})+3=0 \)

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

Substitute:

u = 3x

we have now a Quadratic Equation

u2 - 4*u + 3 = 0

continue....
 

Virgin Vanilla

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\(\displaystyle u2 - 4*u + 3 = 0\)
\(\displaystyle u-3=0\) or \(\displaystyle u-1=0\)
\(\displaystyle u=3\) or \(\displaystyle u=1\\\\\)
if \(\displaystyle u=3\)
\(\displaystyle 3^x=3\)
\(\displaystyle x=1\\\\\)
if \(\displaystyle u=1\)
\(\displaystyle 3^x=1\\
x=1\)
 

Harry_the_cat

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Mar 16, 2016
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Think about your last line again.
 

HallsofIvy

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Can someone please check my answer to the below question?

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

Since the base is same;

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

\(\displaystyle y= 3^x= 3\) 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.
This is simply w
\(\displaystyle 3x=1-1\\
x=0/3\\
x=0\)

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.
 
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