Using Properties of Logarithms

a

We can move the number $2$ down in front of the logarithm using $lo g_{10} (a^{m}) = m \cdot lo g_{10} (a)$ . This gives $lo g (5^{2}) = 2 \cdot lo g (5) .$

b

We'll use the same property and move number $3$ down in front of the logarithm. $lo g (1 1^{3}) = 3 \cdot lo g (11)$

c

Here, there is no exponent in the logarithm, but rewrite

49

as

7^{2}

. Then we use the same property again.

lo g (49)

WritePowWrite as a power

lo g (7^{2})

lo g_{10} (a^{m}) = m \cdot lo g_{10} (a)

2 \cdot lo g (7)

Using Properties of Logarithms 1.17 - Solution