How do I solve $60\cdot 3^{x/8}= 30\cdot 2^{x/5}$ 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. :(
 A: 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  $$x\left(\frac 15\log_{10} 2-\frac 18\log_{10} 3\right)=\log_{10} 2$$ $$\implies x=\frac{40\log_{10} 2}{8\log_{10}2-5\log_{10} 3}≈532\ days.$$
A: 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)
A: 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?
