Proving $x$ is a given quotient of logarithms I'm practicing some questions on logarithms at the moment in order that I'm up to speed with the problem solving aspect before I embark on my PHD in chemical engineering at Boston college next year.
I've been studying the laws of logarithms and what I am to do when it is necessary to add and subtract logs. So for example, for the first part of the question that is causing me trouble, I would utilize the first law of logarithms.
My issue comes with the "Prove part", and ascertaining a value for $x$ when it is already involved in the first part of the question.
I know I'll most likely to be shut down for lack of evidence and research for this, but I would like to know what the most logical first step would be and which rule I need to follow. It doesn't follow on from the other questions I've been solving and so I can't problem solve it as easily as I can the rest.
Here is my question:
If $4^x\cdot 5^{3x+1}=10^{2x+1}$, prove that $x=\dfrac{\log(2)}{\log(5)}$.
Thanks, not a problem if I get shut down, I know this isn't in the spirit of the website and would normally never ask a question in such a manner.
 A: $$4^x\cdot 5^{3x+1} = 10^{2x+1}$$
$~~$Let us begin by moving everything to one side by dividing both sides by $10^{2x+1}$
$$\dfrac{4^x\cdot 5^{3x+1}}{10^{2x+1}} = 1$$
$~~$Let us separate the denominator using the fact that $10=2\cdot 5$
$$\dfrac{4^x\cdot 5^{3x+1}}{(2\cdot 5)^{2x+1}} = 1$$
$~~$Now, $(ab)^c = a^c\cdot b^c$
$$\dfrac{4^x\cdot 5^{3x+1}}{2^{2x+1}\cdot 5^{2x+1}} = 1$$
$~~$Let us factor out a factor of two from the denominator using $a^{b+c} = a^b\cdot a^c$
$$\dfrac{4^x\cdot 5^{3x+1}}{2^{2x}\cdot 2\cdot 5^{2x+1}}=1$$
$~~$Now, using $a^{bc} = (a^b)^c$ simplify $2^{2x}$ in terms of a power of four
$$\dfrac{4^x\cdot 5^{3x+1}}{4^x\cdot 2\cdot 5^{2x+1}}=1$$
$~~$Cancel like terms and use $\frac{a^b}{a^c} = a^{b-c}$
$$\frac{5^x}{2}=1$$
$~~$Multiply both sides by two
$$5^x=2$$
$~~$Take the logarithm of each side
$$\log(5^x)=\log(2)$$
$~~$Use the property of logarithms that $\log(a^b)=b\log(a)$
$$x\log(5)=\log(2)$$
$~~$Divide each side by $\log(5)$ to arrive at the desired result
$$x=\frac{\log(2)}{\log(5)}$$
A: Alternative Solution:
Recall the laws of logarithms:


*

*$\log a^b = b \log a$.

*$\log ab = \log a + \log b$

*$\log \frac{a}{b} = \log a - \log b$


By taking the logarithm on both sides of your equation and applying rule (1), you will get:
$$
x \log 4 + (3x + 1) \log 5 = (2x + 1) \log (10)
$$
which can be simplified to
$$
(\log 4 + 3 \log 5 – 2 \log 10) x = \log(10) - \log(5)
$$
Using the mentioned rules, we get
$$
\log\left(\frac{4 \times 5^3}{10^2}\right) x = \log\left(\frac{10}{5}\right)
$$
which establish your result.
A: \begin{align}
4^x \cdot 5^{3x+1}
    & = 2^{2x} \cdot 5^{3x+1} \\
    & = 2^{2x} \cdot 5^{2x} \cdot 5^{x+1} \\
    & = 10^{2x} \cdot 5^{x+1}
\end{align}
If $10^{2x} \cdot 5^{x+1} = 10^{2x+1}$ (the hypothesis of the problem), then
$$
5^{x+1} = 10
$$
$$
5^x \cdot 5^1 = 10
$$
$$
5^x = 2
$$
which gives us, by definition,
$$
x = \log_5 2
$$
which can be rewritten, using $\log_b x = \frac{\log x}{\log b}$, as
$$
x = \frac{\log 2}{\log 5}
$$
