Question

How do you solve `2^ { m + 1} + 9= 44`?

Answer 1

`m=log_2(35)-1~~4.13`

Explanation:

We start by subtracting `9` from both sides:
`2^(m+1)+cancel(9-9)=44-9`

`2^(m+1)=35`

Take `log_2` on both sides:
`cancel(log_2)(cancel(2)^(m+1))=log_2(35)`

`m+1=log_2(35)`

Subtract `1` on both sides:
`m+cancel(1-1)=log_2(35)-1`

`m=log_2(35)-1~~4.13`

Answer 2

`m~~4.129` (4sf)

Explanation:

`2^(m+1)+9=44`

`2^(m+1)=35`

In logarithm form, this is:

`log_2(35)=m+1`

I remember this almost as keep 2 as the base and switch the other numbers.

`m=log_2(35)-1`

`m~~4.129` (4sf)

Answer 3

`m=(log35-log2)/log2`

Explanation:

`2^(m+1)+9=44`

`2^(m+1)=44-9=35`

`log (2^(m+1))=log35" "` (taking the logarithm base `10` on both sides)

`log(2^m * 2)=log35`

`log2^m +log2=log35`

`log2^m=log35-log2`

`mlog2=log35-log2`

`m=(log35-log2)/log2`