Equation with logarithm in power

[wit]

\(\displaystyle (3x)^{1+\log_3(x)}=3\)

How do I solve this one?

 

[MarkFL]

What do you get if you apply the exponential property:

\(\displaystyle a^b\cdot a^c=a^{b+c}\)?

 

[wit]

Thank you for your reply.

Tried my best, is it any good?

\(\displaystyle (3x)*(3x)^{\log_3(x)}=3\)

\(\displaystyle (3x)*(x^2)=3\) (Not sure about this part)

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

\(\displaystyle (x^3)=1\)

\(\displaystyle (x)=1\)

 

[MarkFL]

Thank you for your reply.

Tried my best, is it any good?

\(\displaystyle (3x)*(3x)^{\log_3(x)}=3\)

\(\displaystyle (3x)*(x^2)=3\) (Not sure about this part)

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

\(\displaystyle (x^3)=1\)

\(\displaystyle (x)=1\)

Your first step is good! :) And kudos on picking up on $\LaTeX$ so quickly! (Yes)

So, we have:

\(\displaystyle 3x(3x)^{\log_3(x)}=3\)

Let's divide through by $3x$ (we know from the original equation $0<x$), so we now have:

\(\displaystyle (3x)^{\log_3(x)}=\frac{1}{x}\)

Now, to the left side, let's apply the exponential rule:

\(\displaystyle (ab)^c=a^cb^c\)

What do we have now?

 

[wit]

Your first step is good! :) And kudos on picking up on $\LaTeX$ so quickly! (Yes)

So, we have:

\(\displaystyle 3x(3x)^{\log_3(x)}=3\)

Let's divide through by $3x$ (we know from the original equation $0<x$), so we now have:

\(\displaystyle (3x)^{\log_3(x)}=\frac{1}{x}\)

Now, to the left side, let's apply the exponential rule:

\(\displaystyle (ab)^c=a^cb^c\)

What do we have now?

\(\displaystyle (3x)^{\log_3(x)}=\frac{1}{x}\)

\(\displaystyle (3)^{\log_3(x)}*(x)^{\log_3(x)}=\frac{1}{x}\) So now I can apply: \(\displaystyle (a)^{\log_a(b)} = b\)?

Which means:

\(\displaystyle (x)*(x)^{\log_3(x)}=\frac{1}{x}\) | /x

\(\displaystyle (x)^{\log_3(x)}=\frac{1}{x^2}\)

And now I think I am confused

 

[MarkFL]

Good job! So, we now have:

\(\displaystyle x^{\log_3(x)}=\frac{1}{x^2}\)

or

\(\displaystyle x^{\log_3(x)}=x^{-2}\)

This implies that the exponents must be the same, right (or that $x=1$, but let's ignore that solution for now)? :)

 

[wit]

Good job! So, we now have:

\(\displaystyle x^{\log_3(x)}=\frac{1}{x^2}\)

or

\(\displaystyle x^{\log_3(x)}=x^{-2}\)

This implies that the exponents must be the same, right (or that $x=1$, but let's ignore that solution for now)? :)

\(\displaystyle \log_3(x)=-2\) | \(\displaystyle \log_a(x)=b => a^{b} = x\)

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

\(\displaystyle x=\frac{1}{9}\)

and as you said $x=1$

Final answer: {\(\displaystyle 1,\frac{1}{9}\)} ... How do I get to \(\displaystyle x=1\), though? It's pretty straight forward that \(\displaystyle 1^{anything} = 1\) but how do I 'prove' it?

 

[MarkFL]

Well, you could go back to:

\(\displaystyle x^{\log_3(x)}=x^{-2}\)

Now, take the natural log of both sides, and apply \(\displaystyle \log_a\left(b^c\right)=c\log_a(b)\) to get:

\(\displaystyle \log_3(x)\ln(x)=-2\ln(x)\)

Then arrange as:

\(\displaystyle \log_3(x)\ln(x)+2\ln(x)=0\)

Now factor:

\(\displaystyle \ln(x)\left(\log_3(x)+2\right)=0\)

Now equate each factor in turn to 0 (apply the zero-factor property) and solve for $x$:

i) \(\displaystyle \ln(x)=0\implies x=1\)

ii) \(\displaystyle \log_3(x)+2=0\implies x=\frac{1}{9}\)

 

[wit]

Been off of math for quite some time now. It seems like even the most basic rules managed to escape my mind.

I cannot be thankful enough for your help. :)

Take care!

 

[Greg]

Alternatively,

\(\displaystyle (3x)^{1+\log_3(x)}=3\)

Take the base 3 log of both sides:

\(\displaystyle (1+\log_3(x))\log_3(3x)=1\)

\(\displaystyle (1+\log_3(x))(1+\log_3(x))=1\) (see note below).

\(\displaystyle (1+\log_3(x))^2=1\)

\(\displaystyle 1+\log_3(x)=\pm1\)

\(\displaystyle \log_3(x)=0\) and \(\displaystyle \log_3(x)=-2\)

\(\displaystyle \log_3(x)=0\implies x=1\)

\(\displaystyle \log_3(x)=-2\)

\(\displaystyle 3^{\log_3(x)}=3^{-2}\)

\(\displaystyle x=\dfrac19\)

Note: \(\displaystyle \log(a\cdot b)=\log a+\log b\) so \(\displaystyle \log_3(3x)=\log_33+\log_3(x)=1+\log_3(x)\)

 

[topsquark]

Been off of math for quite some time now. It seems like even the most basic rules managed to escape my mind.

I cannot be thankful enough for your help. :)

Take care!

Don't be down on yourself. This is not a "trivial" problem for a beginner.

-Dan