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

How do you solve 9*9^(10k-2.7)=339*9^(10k-2.7)=33?