How do I solve $60\cdot 3^{x/8}= 30\cdot 2^{x/5}$

Question

The full question is: Two bacteria cultures are being studied in a lab. At the start, bacteria A had a population of 60 bacteria and the number of bacteria was tripling every 8 days. Bacteria B had a population of 30 bacteria and was doubling every 5 days. Determine the number of days it will take for both bacteria cultures to have the same population.

$$\displaystyle 60\cdot 3^{\tfrac{x}{8}} = 30\cdot 2^{\tfrac{x}{5}}$$ Is the equation I got.

I simplified it to $$2\cdot3^{\tfrac x8} = 2^{\tfrac x5}$$ and got stuck. :(

Please note I have and probably will not be learning LN (log base of e) this course so even if I use it on this question it can't be used on a test. :(

I also find it impossible that you will not learn natural logs in your course if you're given a question like this... — Sean Roberson, Jul 17 '22 at 18:27
Um how do I mark a question as answered so people don't waist time answering an answered question? — Logi boi, Jul 17 '22 at 18:58

insipidintegrator · Accepted Answer · 2022-07-17T18:37:45.043

2

Use the fact that $\displaystyle a=10^{\log_{10} a}$ to get $$10^{\log_{10} 2}\cdot(10^{\log_{10} 3})^{\tfrac x8}=(10^{\log_{10}2})^{\tfrac x5}$$ so that you get $$\displaystyle 10^{\log_{10} 2+\frac x8\log_{10} 3}=10^{\frac x5\log_{10} 2}$$ Since bases are equal, we can equate the exponents to get $$\log_{10} 2+\frac x8\log_{10}3= \frac x5\log_{10}2$$ so that $$x\left(\frac 15\log_{10} 2-\frac 18\log_{10} 3\right)=\log_{10} 2$$ $$\implies x=\frac{40\log_{10} 2}{8\log_{10}2-5\log_{10} 3}≈532\ days.$$

edited Jul 17 '22 at 18:37

answered Jul 17 '22 at 18:23

insipidintegrator

6,642

Sry but I haven't learned LN I know it means log base of Euler's number but I'm sure I can't use it for homework or tests as it was not taught in class sadly. :( – Logi boi Jul 17 '22 at 18:31
Okay I will edit it to $\log_2$. Will that work? Or do you want to put it to $\log_{10}$? – insipidintegrator Jul 17 '22 at 18:33
I’ve edited the post @Logiboi. Now see if you can understand it. – insipidintegrator Jul 17 '22 at 18:38
Edit it to be which ever one would be easier to understand(if it won't be to much work for you). I was able to understand the other homework questions but this one is throwing me off so hard. (I saw ur comment after I posted this one oops). – Logi boi Jul 17 '22 at 18:39
1

Is your question answered? – insipidintegrator Jul 17 '22 at 18:41
nononono it's fine I understand question more now. – Logi boi Jul 17 '22 at 18:41
I should have remembered a=10^{\log_{10} a}$ dumb me :) . – Logi boi Jul 17 '22 at 18:46
Thx for the help guys. – Logi boi Jul 17 '22 at 18:55

score 2 · Answer 2 · answered Jul 17 '22 at 18:47

You can use logs with any base. With base 2 the steps are as follows:

$$60\cdot 3^{\tfrac{x}{8}} = 30\cdot 2^{\tfrac{x}{5}}$$

Taking $\log_2()$ on both sides of the equation.

$$\log_2(60\cdot 3^{\tfrac{x}{8}}) = \log_2(30\cdot 2^{\tfrac{x}{5}})$$

$$\log_2(60)+\log_2( 3^{\tfrac{x}{8}}) = \log_2(30)+\log_2( 2^{\tfrac{x}{5}})$$

$$\log_2(60)+\tfrac{x}{8}\cdot \log_2( 3) = \log_2(30)+\tfrac{x}{5}\cdot \log_2( 2)$$

$\log_2(2)=1$

$$\log_2(60)-\log_2(30)=\tfrac{x}{5}\cdot 1-\tfrac{x}{8}\cdot \log_2( 3)$$

Factoring out $x$

$$\log_2(60)-\log_2(30)=x\cdot \left(\tfrac{1}{5}-\tfrac{1}{8}\cdot \log_2( 3)\right)$$

$$x=\frac{\log_2(60)-\log_2(30)}{\tfrac{1}{5}-\tfrac{1}{8}\cdot \log_2( 3)}=532$$ (rounded to 2 decimal places)

score 0 · Answer 3 · answered Jul 17 '22 at 18:07

0

Make both bases the same: replace the $3$ with $2^{\log_2(3)}$. So then $$2\cdot2^{\log_2(3)\frac{x}{8}}=2^{\frac{x}{5}}$$ So

$$2=2^{\frac{x}{5}-\frac{\log_2(3)}{8}x}$$

Can you take it from here?

answered Jul 17 '22 at 18:07

2'5 9'2

54,717

I will try to thanks. – Logi boi Jul 17 '22 at 18:09
I tried to solve it but my brain started to melt and combust. – Logi boi Jul 17 '22 at 18:32
@Logiboi The next step would be to recognize that $\frac{x}{5}-\frac{\log_2(3)}{8}x=1$. Then you factor out the $x$ to get $x\left(\frac15-\frac{\log_2(3)}{8}\right)=1$. One division and you've isolated $x$. – 2'5 9'2 Jul 18 '22 at 05:33

How do I solve $60\cdot 3^{x/8}= 30\cdot 2^{x/5}$

3 Answers3