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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.

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  • $\begingroup$ For the sake of typesetting, it is recommended to use MathJax and $\LaTeX$, or at the very least use parenthesis so that there is no confusion as to what is meant to go where. Visit this page for information on how to typeset mathematics here. Is the equation meant to be $4^x 5^{3x+1} = 10^{2x+1}$? Is it meant to be $4^{x5^{3x}+1}=10^{2x}+1$ is it meant to be $4^x5^3x+1 = 10^2x+1$ or something else entirely? $\endgroup$ – JMoravitz May 2 '16 at 1:27
  • $\begingroup$ Seems like your question is perfectly in line with the spirit of the website. $\endgroup$ – M47145 May 2 '16 at 1:28
  • $\begingroup$ Thanks for the link, I must learnt the code as it'll prove invaluable to me. The equation is the first you mentioned, with everything being a power apart from the 4, 5 and 10. Thanks, M47145. I'm sat here with an inordinate amount of maths textbooks open trying to get my head around this one! $\endgroup$ – New Zealand's finest May 2 '16 at 1:31
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$$4^x\cdot 5^{3x+1} = 10^{2x+1}$$

$~~$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

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

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    $\begingroup$ It is not recommended to use quotes for highlighting. $\endgroup$ – Ian Miller May 2 '16 at 2:04
  • $\begingroup$ @IanMiller given the amount of text inbetween each step, I find it harder to quickly read without, but have edited it according to your suggestion. Hopefully this is a fine compromise. $\endgroup$ – JMoravitz May 2 '16 at 2:20
  • $\begingroup$ I replaced your single \$ equations with double \$\$ equations to make it look more like your original. $\endgroup$ – Ian Miller May 2 '16 at 2:30
  • $\begingroup$ JMoravitz, thank you for this excellent breakdown. It was invaluable to my learning, honestly. I really appreciate it. $\endgroup$ – New Zealand's finest May 5 '16 at 4:29
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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.

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  • $\begingroup$ Thank you FALAM for helping me to understand this utlizing the laws of logarithms, I really appreciate it. $\endgroup$ – New Zealand's finest May 5 '16 at 4:29
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\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

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

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  • $\begingroup$ Thanks for this version Brian, it was very helpful to me. $\endgroup$ – New Zealand's finest May 5 '16 at 4:30

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