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

2 Answers
Apr 22, 2018

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

Apr 22, 2018

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