MHB Why is the right side of the equation incorrect in 3^(2x) = 18x?

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The equation 3^(2x) = 18x cannot be solved algebraically, but it can be approached using numeric root-finding techniques. By rewriting the equation as 3^(2x) = 18^x, it simplifies to 1 - 2^x = 0, leading to the conclusion that x = 0. The discussion highlights that while logarithms can be used to solve exponential equations, they are not always necessary. Ultimately, the key takeaway is that any non-zero number raised to the power of zero equals one, confirming that x = 0 is the solution.
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Can this be a mistake?

Find x.

3^(2x) = 18x

Shouldn't the right side be 18^(x)?

If so, I can then take the log on both sides.
 
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Yes, the equation as given cannot be solved algebraically. A numeric root-finding technique could be used though. If we take the equation to be:

$$3^{2x}=18^x$$

Then we could write this as:

$$3^{2x}-2^x\cdot3^{2x}=0$$

Divide through by $3^{2x}$ since it cannot be zero for real values of $x$:

$$1-2^x=0\implies x=0$$

You could also use logs, but I just wanted to present an alternate method. :D
 
Is this not an exponential equation?
 
RTCNTC said:
Is this not an exponential equation?

Yes, but it's not always necessary to use logs to solve them. If we can arrange such an equation in the form:

$$a^b=a^c$$ where $a$ is neither 0 nor 1, then we know:

$$b=c$$

In this problem, we found:

$$2^x=1=2^0$$

Therefore we conclude:

$$x=0$$
 
MarkFL said:
Yes, the equation as given cannot be solved algebraically. A numeric root-finding technique could be used though. If we take the equation to be:

$$3^{2x}=18^x$$

Then we could write this as:

$$3^{2x}-2^x\cdot3^{2x}=0$$

Divide through by $3^{2x}$ since it cannot be zero for real values of $x$:

$$1-2^x=0\implies x=0$$

You could also use logs, but I just wanted to present an alternate method. :D

3^2x = 18^x

3^2x - 2^x • 3^2x = 0

You said "Divide through by 3^2x..."

We are left with 1 - 2^x = 0.

1 = 2^x

ln(1) = ln[2^(x)]

0 = xln2

0/ln2 = x

0 = x
 
Once you reach the point:

$$2^x=1$$

Instead of taking the natural log of both sides (which is completely valid), you could also translate directly from exponential to logarithmic notation, and write:

$$x=\log_2(1)=0$$ :D
 
MarkFL said:
Once you reach the point:

$$2^x=1$$

Instead of taking the natural log of both sides (which is completely valid), you could also translate directly from exponential to logarithmic notation, and write:

$$x=\log_2(1)=0$$ :D

We can also say, in terms of 2^x = 1, that anything to the zero power is 1.

So, 2^x = 1 means (number)^0 = 1.

We say x = 0.
 
RTCNTC said:
We can also say, in terms of 2^x = 1, that anything to the zero power is 1.

So, 2^x = 1 means (number)^0 = 1.

We say x = 0.

Yes, any non-zero value raised to the power of zero is 1, as evidenced by using a rule for exponents:

$$1=\frac{a^b}{a^b}=a^{b-b}=a^0$$
 
MarkFL said:
Yes, any non-zero value raised to the power of zero is 1, as evidenced by using a rule for exponents:

$$1=\frac{a^b}{a^b}=a^{b-b}=a^0$$

Very cool.
 

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