# Thread: check work

1. ## 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

2. 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 $\log_5(x+2)=\log_5 25\implies\dots\implies x=\dots$

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?

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...)

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?

3. 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 $2^{x+2}$

4. 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:
$2^x - 4 \times 2^x = -3 \times 2^x$

5. 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

6. 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.

7. 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

8. 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 ^.^