For a>0, a 1 and N>0, log_a N=u means au=N. The base is a positive number other than 1; the argument must be positive. The logarithm asks which index produces the argument. Powers with negative or fractional indices give logarithms with negative or fractional values: 4- 32= 18, so log_4!(18)=- 32. Method: identify the base and argument; rewrite the statement as a power; express both sides in one base where possible; check the original argument. Work from the outside inward for nested logarithms. For log_2(log_3 x)=1, first log_3 x=2, then x=9. Both original arguments are positive. Exam check: log_a 1=0, but log_a 0 is undefined. A base between 0 and 1 is allowed; a negative base is not a real logarithm base here.

For \(a>0\), \(a\ne1\) and \(N>0\), \(\log_a N=u\) means \(a^u=N\). The base is a positive number other than \(1\); the argument must be positive.

The logarithm asks which index produces the argument. Powers with negative or fractional indices give logarithms with negative or fractional values: \(4^{\frac{-3}{2}}=\dfrac18\), so \(\log_4\!\left(\dfrac18\right)=-\dfrac32\).

Method: identify the base and argument; rewrite the statement as a power; express both sides in one base where possible; check the original argument. Work from the outside inward for nested logarithms.

For \(\log_2(\log_3x)=1\), first \(\log_3x=2\), then \(x=9\). Both original arguments are positive.

Exam check: \(\log_a1=0\), but \(\log_a0\) is undefined. A base between \(0\) and \(1\) is allowed; a negative base is not a real logarithm base here.

Logarithms as indices

Logarithms as indices