Question

How do you solve `2^(n+4)=1/32`?

Answer 1

`n=(-ln32/ln2)-4`

Explanation:

Apply the natural logarithm to both sides:

`ln(2^(n+4))=ln(1/32)`

Recall the logarithm quotient rule, which tells us that `ln(a/b)=lna-lnb.` Then, `ln(1/32)=ln1-ln32,` and as `ln1=0,` we get `-ln32.`

Furthermore, recall the logarithm exponent rule: `ln(a^b)=blna.` So, `ln(2^(n+4))=(n+4)ln2`, and we end up with

`(n+4)ln2=-ln32`

Solve for `n`:

`(n+4)=-ln32/ln2`

`n=(-ln32/ln2)-4`

Answer 2

`n=-9`

Explanation:

`"note that "1/32=1/2^5=2^-5`

`rArr2^(n+4)=2^-5larrcolor(blue)"equation to be solved"`

`"Since the bases on both sides of the equation are equal"`
`"that is both 2, we can equate the exponents"`

`rArrn+4=-5rArrn=-9`