we'll use the same log cancellation rule, since this is pretty much the same thing as the other, just recall that ln = logₑ.
![\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{a^{log_a x}=x} \\\\\\ \textit{logarithm of factors} \\\\ log_a(xy)\implies log_a(x)+log_a(y) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ln(x)+ln(5)=2\implies ln(x\cdot 5)=2\implies log_e(5x)=2 \\\\\\ e^{log_e(5x)}=e^2\implies 5x=e^2\implies x=\cfrac{e^2}{5}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7BLogarithm%20Cancellation%20Rules%7D%0A%5C%5C%5C%5C%0Alog_a%20a%5Ex%20%3D%20x%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba%5E%7Blog_a%20x%7D%3Dx%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Blogarithm%20of%20factors%7D%0A%5C%5C%5C%5C%0Alog_a%28xy%29%5Cimplies%20log_a%28x%29%2Blog_a%28y%29%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0Aln%28x%29%2Bln%285%29%3D2%5Cimplies%20ln%28x%5Ccdot%205%29%3D2%5Cimplies%20log_e%285x%29%3D2%0A%5C%5C%5C%5C%5C%5C%0Ae%5E%7Blog_e%285x%29%7D%3De%5E2%5Cimplies%205x%3De%5E2%5Cimplies%20x%3D%5Ccfrac%7Be%5E2%7D%7B5%7D%20)
![\bf \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ln(x)-ln(5)=2\implies ln\left( \cfrac{x}{5} \right)=2\implies log_e\left( \cfrac{x}{5} \right)=2\implies e^{log_e\left( \frac{x}{5} \right)}=e^2 \\\\\\ \cfrac{x}{5}=e^2\implies x=5e^2](https://tex.z-dn.net/?f=%20%5Cbf%20%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0Aln%28x%29-ln%285%29%3D2%5Cimplies%20ln%5Cleft%28%20%5Ccfrac%7Bx%7D%7B5%7D%20%5Cright%29%3D2%5Cimplies%20log_e%5Cleft%28%20%5Ccfrac%7Bx%7D%7B5%7D%20%5Cright%29%3D2%5Cimplies%20e%5E%7Blog_e%5Cleft%28%20%5Cfrac%7Bx%7D%7B5%7D%20%5Cright%29%7D%3De%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bx%7D%7B5%7D%3De%5E2%5Cimplies%20x%3D5e%5E2%20)
Answer:
6/5 or 1 and 1/5
Step-by-step explanation:
To do this you need to multiply 6/1, 2/3, and 3/10 all together by cancelling out the numerator in 3/10 and the denominator for 2/3. Then you would be left with 6*2*1/10
This simplifies to 6*1/5 because of the 2*1/10 is 1/5, because 1/5 doubled is 1/10.
then multiply the numerator and 6 together to get 6/5.
Hope this helps!
Okay well
a(t)= amount of substance remaining after t days
assuming exponential decay, a(t)=29e^-0.1359t
now, find the half, set a(t)= half the original and solve for t
14.5=29e^-0.1359t
0.5=e^-0.1359t
In(0.5)=In(e^-0.1359t)
In(0.5)=-o.1359t
t=in(0.5)/(-0.1359)= (aprox) 5.1 days
Answer:
variables
Step-by-step explanation:
Are you asking for an answer like this?
Answer:
line graph for sure ig..........
