Answer:
When dividing both sides by a negative.
Step-by-step explanation:
When dividing OR multiplying both sides by a negative, we need to flip the inequality signs.
Answer:
80, 95, 110, 125, 140
Step-by-step explanation:
<u>Given</u>:
The given expression is ![2^{x+5}=13^{2 x}](https://tex.z-dn.net/?f=2%5E%7Bx%2B5%7D%3D13%5E%7B2%20x%7D)
We need to determine the value of x using either base - 10 or base - e logarithms.
<u>Value of x:</u>
Let us determine the value of x using the base - e logarithms.
Applying the log rule that if
then ![\ln (f(x))=\ln (g(x))](https://tex.z-dn.net/?f=%5Cln%20%28f%28x%29%29%3D%5Cln%20%28g%28x%29%29)
Thus, we get;
![\ln \left(2^{x+5}\right)=\ln \left(13^{2 x}\right)](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%282%5E%7Bx%2B5%7D%5Cright%29%3D%5Cln%20%5Cleft%2813%5E%7B2%20x%7D%5Cright%29)
Applying the log rule,
, we get;
![(x+5) \ln (2)=2 x \ln (13)](https://tex.z-dn.net/?f=%28x%2B5%29%20%5Cln%20%282%29%3D2%20x%20%5Cln%20%2813%29)
Expanding, we get;
![x \ln (2)+5 \ln (2)=2 x \ln (13)](https://tex.z-dn.net/?f=x%20%5Cln%20%282%29%2B5%20%5Cln%20%282%29%3D2%20x%20%5Cln%20%2813%29)
Subtracting both sides by
, we get;
![x \ln (2)=2 x \ln (13)-5 \ln (2)](https://tex.z-dn.net/?f=x%20%5Cln%20%282%29%3D2%20x%20%5Cln%20%2813%29-5%20%5Cln%20%282%29)
Subtracting both sides by
, we get;
![x \ln (2)-2 x \ln (13)=-5 \ln (2)](https://tex.z-dn.net/?f=x%20%5Cln%20%282%29-2%20x%20%5Cln%20%2813%29%3D-5%20%5Cln%20%282%29)
Taking out the common term x, we have;
![x( \ln (2)-2 \ln (13))=-5 \ln (2)](https://tex.z-dn.net/?f=x%28%20%5Cln%20%282%29-2%20%5Cln%20%2813%29%29%3D-5%20%5Cln%20%282%29)
![x=\frac{-5 \ln (2)}{\ln (2)-2 \ln (13)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-5%20%5Cln%20%282%29%7D%7B%5Cln%20%282%29-2%20%5Cln%20%2813%29%7D)
![x=\frac{5 \ln (2)}{2 \ln (13)-\ln (2)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B5%20%5Cln%20%282%29%7D%7B2%20%5Cln%20%2813%29-%5Cln%20%282%29%7D)
Thus, the value of x is ![x=\frac{5 \ln (2)}{2 \ln (13)-\ln (2)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B5%20%5Cln%20%282%29%7D%7B2%20%5Cln%20%2813%29-%5Cln%20%282%29%7D)
<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>
Step-by-step explanation:
Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.
From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by
=
.
So, the population in the year t can be given by ![P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}](https://tex.z-dn.net/?f=P%28t%29%3D3%2C381%2C000%5Ctextrm%7Bx%7D%28%5Cfrac%7B101.8%7D%7B100%7D%29%5E%7B%28t-1994%29%7D)
Population in the year 2000 =
=![3,762,979.38](https://tex.z-dn.net/?f=3%2C762%2C979.38)
Population in year 2000 = 3,762,979
Let us assume population doubles by year
.
![2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}](https://tex.z-dn.net/?f=2%5Ctextrm%7Bx%7D%283%2C381%2C000%29%3D%283%2C381%2C000%29%5Ctextrm%7Bx%7D%28%5Cfrac%7B101.8%7D%7B100%7D%29%5E%7B%28y-1994%29%7D)
![log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})](https://tex.z-dn.net/?f=log_%7B10%7D2%3D%28y-1994%29log_%7B10%7D%28%5Cfrac%7B101.8%7D%7B100%7D%29)
![y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537](https://tex.z-dn.net/?f=y-1994%3D%5Cfrac%7Blog_%7B10%7D2%7D%7Blog_%7B10%7D1.018%7D%3D38.8537)
≈![2033](https://tex.z-dn.net/?f=2033)
∴ By 2033, the population doubles.
Answer:
![3/7](https://tex.z-dn.net/?f=3%2F7)
Step-by-step explanation:
![3/7 \approx 0.42857142857](https://tex.z-dn.net/?f=3%2F7%20%5Capprox%200.42857142857)
![1/3 \approx 0.33333333333](https://tex.z-dn.net/?f=1%2F3%20%5Capprox%200.33333333333)
![2/5 = 0.4](https://tex.z-dn.net/?f=2%2F5%20%3D%200.4)
![3/8 = 0.375](https://tex.z-dn.net/?f=3%2F8%20%3D%200.375)