Logarithmic equation $\log n\le10^6$

Question

I'm trying to solve logarithmic equation (instance below).

$\log{n}\le10^6$

Solution is:

$\log{n}\le10^6\rightarrow2^{\log{n}}\le2^{10^6}\rightarrow n\le2^{10^6}$.

I have no idea how author get this solution. This mean there have a property like $x^{\log{n}}=n$? And how solve this with using real log properties?

Source Row 1 https://udel.edu/~caviness/Class/CISC320-02S/exercise-solns/ch01/R-1.7.pdf

Answer 1

$$\log(n) \le 10^6$$

From the elementary property of the exponential-logarithm we have that

$$\large b^{\log_b(x)} = x ~~~~~~~~~~~ \text{for} ~~~ x > 0;\ ~~ b > 0,\ ~~ b\neq 1$$

From what I understand, your "book" or whatever, uses the ugly notation $\log$ to indicate $\log_2$, the binary logarithm, that is, base $2$.

Besides the ugliness, we exponentiate in base $2$ then:

$$\large 2^{\log_2(n)} \leq 2^{10^6}$$

$$n \leq 2^{10^6}$$

Which provides the result you were searching for.

Answer 2

$$\log_a(n)<10^6$$
But we can use that $10^6=\log_a\left(a^{10^6}\right)$: $$\log_a(n)<\log_a\left(a^{10^6}\right)$$ For $a>1$ the $\log_a$ is increasing, so: $$n<a^{10^6}$$ For $a<1$ the $\log_a$ is decreasing, so: $$n>a^{10^6}$$

Answer 3

Note that by definition

$$c=\log_a b \iff a^c=b$$

thus

$$a^{\log_a b}=b$$

morover since logarithmic function is an increasing function we have that

$$\log_a x\le \log_a y \iff x \le y$$

thus

$$\log_2 n\le 10^6 = \log_2 2^{10^6} \iff n \le 2^{10^6}$$