How do you solve #9*9^(10k-2.7)=33#?

1 Answer
Aug 3, 2016

Answer:

#k = 0.33# to 2 dp

Explanation:

Remember how we multiply exponents:

#x^a*x^b = x^(a+b)#

We have #9^1*9^(10k-2.7) = 33#

#therefore 9^(10k-1.7) = 33#

Take natural logs of both sides and use rules of logs to bring exponent down.

#(10k-1.7)ln(9) = ln(33)#

#10k-1.7 = (ln(33))/(ln(9))#

#k = 1/10((ln(33))/(ln(9)) + 1.7)#

#k = 0.33# to 2 dp