Equation with logarithm in power

[wit]

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

How do I solve this one?

 

[MarkFL]

What do you get if you apply the exponential property:

$$a^b\cdot a^c=a^{b+c}$$?

 

[wit]

Thank you for your reply.

Tried my best, is it any good?

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

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

$$(3x^3)=3$$

$$(x^3)=1$$

$$(x)=1$$

 

[MarkFL]

Thank you for your reply.

Tried my best, is it any good?

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

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

$$(3x^3)=3$$

$$(x^3)=1$$

$$(x)=1$$

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

So, we have:

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

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

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

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

$$(ab)^c=a^cb^c$$

What do we have now?

 

[wit]

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

So, we have:

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

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

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

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

$$(ab)^c=a^cb^c$$

What do we have now?

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

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

Which means:

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

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

And now I think I am confused

 

[MarkFL]

Good job! So, we now have:

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

or

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

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:

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

or

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

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

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

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

$$x=\frac{1}{9}$$

and as you said $x=1$

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

 

[MarkFL]

Well, you could go back to:

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

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

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

Then arrange as:

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

Now factor:

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

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

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

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

 

[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,

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

Take the base 3 log of both sides:

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

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

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

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

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

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

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

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

$$x=\dfrac19$$

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

 

[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