Proving $x$ is a given quotient of logarithms

Question

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.

Answer 1

$$4^x\cdot 5^{3x+1} = 10^{2x+1}$$

$~~$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

$$x=\frac{\log(2)}{\log(5)}$$

Answer 2

Alternative Solution:

Recall the laws of logarithms:

1. $\log a^b = b \log a$.
2. $\log ab = \log a + \log b$
3. $\log \frac{a}{b} = \log a - \log b$

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.

Answer 3

\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

$$ x = \frac{\log 2}{\log 5} $$