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

[mathdad]

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.

 

[MarkFL]

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:

\(\displaystyle 3^{2x}=18^x\)

Then we could write this as:

\(\displaystyle 3^{2x}-2^x\cdot3^{2x}=0\)

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

\(\displaystyle 1-2^x=0\implies x=0\)

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

 

[mathdad]

Is this not an exponential equation?

 

[MarkFL]

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:

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

\(\displaystyle b=c\)

In this problem, we found:

\(\displaystyle 2^x=1=2^0\)

Therefore we conclude:

\(\displaystyle x=0\)

 

[mathdad]

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:

\(\displaystyle 3^{2x}=18^x\)

Then we could write this as:

\(\displaystyle 3^{2x}-2^x\cdot3^{2x}=0\)

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

\(\displaystyle 1-2^x=0\implies x=0\)

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

 

[MarkFL]

Once you reach the point:

\(\displaystyle 2^x=1\)

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:

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

 

[mathdad]

Once you reach the point:

\(\displaystyle 2^x=1\)

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:

\(\displaystyle x=\log_2(1)=0\) :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.

 

[MarkFL]

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:

\(\displaystyle 1=\frac{a^b}{a^b}=a^{b-b}=a^0\)

 

[mathdad]

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

\(\displaystyle 1=\frac{a^b}{a^b}=a^{b-b}=a^0\)

Very cool.