Solving $15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1$

Question

I need help solving this logarithm exercise:

$$15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1$$

What I've done is re-writing the equation

$$\Rightarrow \qquad 5^{\log_5(3)}\cdot 3^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1 \tag{1}$$

Then applying logarithms on both sides

$$\Rightarrow \qquad \log_5(3^{\log_5(3)}\cdot5^{\log_5(3)}\cdot x^{log_5(9x)+1})=\log_5(1) \tag{2}$$

re-writing the equation a little bit

$$\Rightarrow \qquad \log_53+\log_53^{\log_5(3)}+\log_5x^{\log_5(9x)+1}=\log_55 \tag{3}$$

But then I'm not entirely sure how to proceed

Answer 1

I use the change of base to get in terms of $e$ and $\ln$ and make this much easier.

$$\log_a b=\frac{\ln b}{\ln a} \text{ and } p^q=e^{q\ln p}$$

We get: $\log_5 3=\frac{\ln 3}{\ln 5}$, $\log_5(9x+1)=\frac{\ln(9x+1)}{\ln 5}$ and reduce our equation to:

$$e^{\frac{\ln(15)\ln(3)+\ln(x)\ln(9x+1)}{\ln 5}}=1$$

So we solve:

$$\frac{\ln(15)\ln(3)+\ln(x)\ln(9x+1)}{\ln 5}=0$$

Answer 2

I doubt that there are solutions. Observe that if $x=15,$ then necessarily we should have $\log_5(9x)+1=-\log_53.$ But these are inconsistent.

Answer 3

That last edit makes it possible. $\log_5(9x)+1=\log_5(3)+\log_5(3)+\log_5(x)+\log_5(5)=\log_5(x)+\log_5(3)+\log_5(15)$. then on takings logarithms base $5$, $$\begin{align}\left(\log_5(15)\right)\left(\log_5(3)\right)+\left(\log_5(x)\right)\left(\log_5(x)+\log_5(3)+\log_5(15)\right)\\ =\left((\log_5(x)+\log_5(3)\right)\left(\log_5(x)+\log_5(15)\right)=0\end{align}$$ Thus either $\log_5(x)=-\log_5(3)$, so $x=1/3$, or $\log_5(x)=-\log_5(15)$ and $x=1/15$.

Answer 4

First, note that $\log_5(9x)+1 = \log_5(9x)+\log_5(5) = \log_5(45x)$:

$$15^{\log_5(3)}\cdot x^{\log_5(9x)+1} = 1 \iff 15^{\log_5(3)}\cdot x^{\log_5(45x)} = 1$$

Now, isolating the factor with $x$'s, you get

$$x^{\log_5(45x)} = \frac{1}{15^{\log_5(3)}} = 15^{-\log_5(3)} = 15^{\log_5\left(\frac{1}{3}\right)}$$

$$x^{\log_5(45x)} = 15^{\log_5\left(\frac{1}{3}\right)}$$

The first solution is found by comparing both sides. This gives $x = \dfrac{1}{15}$.

For the second solution, notice that $a^{\log_a(c)} = c$, so by changing bases, $a^{\frac{\log_b(c)}{\log_b(a)}} = c \iff a^{\log_b(c)} = c^{\log_b(a)}$. This means that $15^{\log_5\left(\frac{1}{3}\right)} = \left(\frac{1}{3}\right)^{\log_5(15)}$. Hence,

$$x^{\log_5(45x)} = \left(\frac{1}{3}\right)^{\log_5(15)}$$

This gives $x = \dfrac{1}{3}$. You could've also re-written the LHS rather than the RHS, but that would give the same result anyway.

Answer 5

$15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1\\ x^{\log_5(9x)+1}=15^{-\log_5(3)}$

Take the $\log_5$ of both sides

$(\log_5(9x)+1)\log_5 x =(-\log_5(3))(\log_5 15)\\ (\log_5 x + \log_5 9+1)\log_5 x =(-\log_5(3))(\log_5 15)\\ (\log_5 x)^2 + (\log_5 9+1)\log_5 x + \log_5(3)(\log_5 3 + 1) = 0\\ $

let $u = \log_5 x, b = 2\log_5 3 + 1, c = (\log_5 3)^2 + \log_5 3$

$u^2 + bu + c = 0$

$u = \frac {-b \pm \sqrt {b^2 - 4c}}{2}$

$b^2 - 4c = (2\log 3 + 1)^2 - 4(\log^2 3 + \log 3)\\ 4\log^2 3 + 4\log 3 + 1 - 4\log^2 3 - 4\log 3 = 1$

$u = -\log_5 3, -\log_5 3 - 1\\ x = 5^u$

$x = \frac {1}{3}$ or $\frac {1}{15}$

Answer 6

Let $t=\log_5x$ and $a=\log_53$ to rewrite the given equation as

$$t^2+(1+2a)t + (a+a^2) = 0$$ which factorizes $(t+a)(t+1+a)=0$ and yields $t = -a$ and $t=-1-a$, hence $$x=\frac13,\>\frac1{15}$$