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

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Answer 1 · 2016-08-03T07:35:46

#k = 0.33# to 2 dp

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