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#