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log x^3 =
4^2 = 16
5^3 = 125
3/9 = 1/3
11/6
logx - logy
37/30
3logx
y = k/x
y varies inversely with x
4^2 = 16
5^3 = 125
3/9 = 1/3
11/6
logx - logy
37/30
3logx
y = k/x
7/9 - 4/9
4^2 = 16
5^3 = 125
3/9 = 1/3
11/6
logx - logy
37/30
3logx
y = k/x
log(x/y) =
4^2 = 16
5^3 = 125
3/9 = 1/3
11/6
logx - logy
37/30
3logx
y = k/x
Write in exponential form: log base 4 of 16 = 2
4^2 = 16
5^3 = 125
3/9 = 1/3
11/6
logx - logy
37/30
3logx
y = k/x
Write in exponential form: log base 5 of 125 = 3
4^2 = 16
5^3 = 125
3/9 = 1/3
11/6
logx - logy
37/30
3logx
y = k/x
2/5 + 5/6
4^2 = 16
5^3 = 125
3/9 = 1/3
11/6
logx - logy
37/30
3logx
y = k/x
3/2 + 1/3
4^2 = 16
5^3 = 125
3/9 = 1/3
11/6
logx - logy
37/30
3logx
y = k/x
Check it!