# check work2

• Jul 20th 2009, 09:10 AM
william
check work2
1. if $\displaystyle 3^{2x-3}=27$

x=2

2. if $\displaystyle 2^{x-1}= \frac{1}{16}$

x=5

3. if $\displaystyle 2^{x+2}-2^x=12$

x=-2

4. if $\displaystyle 300(1.05)^n=600$

n=2.78

5. to solve $\displaystyle 3^{x+2}-3^x=216$ you must use the logarithmic method
• Jul 20th 2009, 10:43 AM
earboth
Quote:

Originally Posted by william
1. if $\displaystyle 3^{2x-3}=27$

x=2 <<< No

2. if $\displaystyle 2^{x-1}= \frac{1}{16}$

x=5 <<< No

3. if $\displaystyle 2^{x+2}-2^x=12$

x=-2 <<< No

4. if $\displaystyle 300(1.05)^n=600$

n=2.78 <<< No

5. to solve $\displaystyle 3^{x+2}-3^x=216$ you must use the logarithmic method

How did you do these calculations?

Obviously you need a lot of help. But we must know where and why you've made the mistakes. Otherwise we can't help you sufficiently.

I'll take #5:

$\displaystyle 3^{x+2}-3^x=216~\implies~9\cdot 3^{x}-3^x=216 ~\implies~8 \cdot 3^x=216 ~\implies~3^x=27~\implies~x=3$
• Jul 21st 2009, 11:21 AM
william
Quote:

Originally Posted by earboth
How did you do these calculations?

Obviously you need a lot of help. But we must know where and why you've made the mistakes. Otherwise we can't help you sufficiently.

I'll take #5:

$\displaystyle 3^{x+2}-3^x=216~\implies~9\cdot 3^{x}-3^x=216 ~\implies~8 \cdot 3^x=216 ~\implies~3^x=27~\implies~x=3$

for 1. x=2
2. x=-3
3. x=2
4. n=0.30
• Jul 21st 2009, 11:34 AM
e^(i*pi)
Quote:

Originally Posted by william
1. if $\displaystyle 3^{2x-3}=27$

x=2

2. if $\displaystyle 2^{x-1}= \frac{1}{16}$

x=5

3. if $\displaystyle 2^{x+2}-2^x=12$

x=-2

4. if $\displaystyle 300(1.05)^n=600$

n=2.78

5. to solve $\displaystyle 3^{x+2}-3^x=216$ you must use the logarithmic method

Quote:

Originally Posted by william

for 1. x=2
2. x=-3
3. x=2
4. n=0.30

No, x still isn't 2.
In this case note that 27 = 3^3

2. Yes

3. Yup

4. nay

Divide both sides through by 300:

$\displaystyle 1.05^n = 2$
• Jul 21st 2009, 12:06 PM
earboth
Quote:

Originally Posted by william

for 1. x=2
...

You can check yourself if you've got the correct solution:

Plug in the value of x into the original equation. If you get the same value on both sides of the equation your solution is OK. If not then ... (Crying)

$\displaystyle 3^{2x-3}=27$ plug in x = 2 (that's your solution!):

$\displaystyle 3^{2\cdot 2-3}=27~\implies~3^1 = 27~\implies~3 \neq 27$

• Jul 21st 2009, 12:09 PM
william
Quote:

Originally Posted by e^(i*pi)
No, x still isn't 2.
In this case note that 27 = 3^3

2. Yes

3. Yup

4. nay

Divide both sides through by 300:

$\displaystyle 1.05^n = 2$

so 1. x=3

4. x=14.21
• Jul 21st 2009, 12:14 PM
e^(i*pi)
Quote:

Originally Posted by william
so 1. x=3

4. x=14.21

To check do what earboth suggested in post 5 but I get those from my calculations