Using Properties of Logarithms

When bases are not the same, we can solve an exponential equation by taking logarithms on each side of the equation.

m = n \Leftrightarrow lo g (m) = lo g (n)

Note that in order to take their logarithms, both

m

and

n

must be positive numbers. Let's now solve our equation.

6^{3 x - 9} - 10 = - 3

LHS + 10 = RHS + 10

6^{3 x - 9} = 7

lo g (LHS) = lo g (RHS)

lo g (6^{3 x - 9}) = lo g (7)

lo g (a^{m}) = m \cdot lo g (a)

(3 x - 9) lo g (6) = lo g (7)

LHS / lo g (6) = RHS / lo g (6)

3 x - 9 = \frac{lo g ( 7 )}{lo g ( 6 )}

lo g_{c} a = \frac{lo g _{b} a}{lo g _{b} c}

3 x - 9 = lo g_{6} (7)

LHS + 9 = RHS + 9

3 x = lo g_{6} (7) + 9

LHS / 3 = RHS / 3

x = \frac{1}{3} lo g_{6} (7) + 3

The exact solution is

x = \frac{1}{3} lo g_{6} (7) + 3 .

To find an approximate solution to three decimal places, we will use a calculator.

x = \frac{1}{3} lo g_{6} (7) + 3 Use a calculator x \approx 3.362

Using Properties of Logarithms 1.3 - Solution