# check work

• Jul 20th 2009, 09:05 AM
william
check work
1. the first step to solve the equation $log_5(x+2)=2$ is to take logs with base 2 of both sides

2. the first step in solving the equation $2^x-2^{x+2}=8$ is divide all terms by 2

3. the first error in the solved equation
$\begin{array}{l}log(x+5)=2\\
log(x+5)=log2\\
x+5=2\\
x=7\end{array}$

is taking the log of 2

4. the next step in the partially solved equation
$\begin{array}{l}
30(1.02)^n=600\\
(1.02)^n=20\end{array}$

is bring n infront of the 1.02
• Jul 20th 2009, 09:34 AM
Chris L T521
Quote:

Originally Posted by william
1. the first step to solve the equation $log_5(x+2)=2$ is to take logs with base 2 of both sides

Hint:
Spoiler:
$2=2\cdot 1 =2\log_5 5=\log_5 25$
So $\log_5(x+2)=\log_5 25\implies\dots\implies x=\dots$

Quote:

2. the first step in solving the equation $2^x-2^{x+2}=8$ is divide all terms by 2
No. Note that $2^{x+2}=2^2\cdot 2^x=4\cdot 2^x$. Thus $2^x-2^{x+2}=2^x-2\cdot 2^x=-3\cdot 2^x=8$. Do you think this has a solution?

Quote:

3. the first error in the solved equation
$\begin{array}{l}log(x+5)=2\\
log(x+5)=log2\\
x+5=2\\
x=7\end{array}$

is taking the log of 2
Very good. What would be the correct first step? (See my spoiler response to your first problem and follow the same idea...)

Quote:

4. the next step in the partially solved equation
$\begin{array}{l}
30(1.02)^n=600\\
(1.02)^n=20\end{array}$

is bring n infront of the 1.02
Yes, but how do you bring the n out in front?
• Jul 21st 2009, 11:24 AM
william
so for 1. would separation of $log_5(x+2) to log_5x+log_52$ work?

and 2. combine $2^x-2^{x+2}$ into $2^{x+2}$
• Jul 21st 2009, 11:31 AM
e^(i*pi)
Quote:

Originally Posted by william
so for 1. would separation of $log_5(x+2) to log_5x+log_52$ work?

and 2. combine $2^x-2^{x+2}$ into $2^{x+2}$

1&2. No it would not

You can only use those laws when the terms are multiplied or divided

1 is fine as is

2 could be put as:
$2^x - 4 \times 2^x = -3 \times 2^x$
• Jul 21st 2009, 12:14 PM
william
Quote:

Originally Posted by e^(i*pi)
1&2. No it would not

You can only use those laws when the terms are multiplied or divided

1 is fine as is

2 could be put as:
$2^x - 4 \times 2^x = -3 \times 2^x$

for 1 can you put it into exponential form?

for 2 could you write 8 as 2^3 or separate (2)^x(2)^2
• Jul 21st 2009, 12:19 PM
e^(i*pi)
Quote:

Originally Posted by william
for 1 can you put it into exponential form?

for 2 could you write 8 as 2^3 or separate (2)^x(2)^2

1. If you like. If $f(x) = log_5{(x+2)}$ then $x+2 = 5^{f(x)}$ and $x = 5^{f(x)} - 2$

2. You can write 8 as 2^3. I broke down $2^{x+2}$ into $2^22^x$, it is just more usual to simplify 2^2 to 4.
• Jul 21st 2009, 12:46 PM
william
Quote:

Originally Posted by e^(i*pi)
1. If you like. If $f(x) = log_5{(x+2)}$ then $x+2 = 5^{f(x)}$ and $x = 5^{f(x)} - 2$

2. You can write 8 as 2^3. I broke down $2^{x+2}$ into $2^22^x$, it is just more usual to simplify 2^2 to 4.

for 2, which step would come first? writing 8 as 2^3 or breaking 2^{x+2} down into 2^22^x
• Jul 21st 2009, 01:04 PM
e^(i*pi)
Quote:

Originally Posted by william
for 2, which step would come first? writing 8 as 2^3 or breaking 2^{x+2} down into 2^22^x

It doesn't matter ^.^