#10; #10;To solve: #160; log_xx We know that log_a b=log blog a #160; #8230; #8230; (1) ∴log_xx=log xlog x=1 ...

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Theorems for Continuity

Solve: log ...

Question

Solve: ${log}_{x} x$

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Solution

To solve: ${log}_{x} x$
We know that ${log}_{a} b = \frac{log b}{log a}$ …… (1)
$∴ {log}_{x} x = \frac{log x}{log x} = 1$

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