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

Aug 3, 2016

$k = 0.33$ to 2 dp

#### Explanation:

Remember how we multiply exponents:

${x}^{a} \cdot {x}^{b} = {x}^{a + b}$

We have ${9}^{1} \cdot {9}^{10 k - 2.7} = 33$

$\therefore {9}^{10 k - 1.7} = 33$

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

$\left(10 k - 1.7\right) \ln \left(9\right) = \ln \left(33\right)$

$10 k - 1.7 = \frac{\ln \left(33\right)}{\ln \left(9\right)}$

$k = \frac{1}{10} \left(\frac{\ln \left(33\right)}{\ln \left(9\right)} + 1.7\right)$

$k = 0.33$ to 2 dp