Exponential and Logarithmic Function

Which symbol denotes the set of real numbers?

The set $N$ denotes which of the following?

Which symbol represents the set of integers?

The set $Q$ denotes which of the following?

In the expression $a^n$, what is $n$ called?

In $3^4$, what is the base?

In $\left(\dfrac{2}{5}\right)^4$, what is the base?

By definition, $a^1$ equals which of the following?

When the exponent is irrational, the base $a$ must satisfy which condition?

Approximate value of $\sqrt{5}$ used in computing $3^{\sqrt{5}}$ is:

Approximate value of $3^{\sqrt{5}}$ is:

Logarithm value of which type of number can NOT be determined?

According to Formula 1, $a^{n+1}$ equals:

The functional law of exponents states $a^m \cdot a^n =$

For $m > n$ and $a \neq 0$, $\dfrac{a^m}{a^n}$ equals:

For $m < n$ and $a \neq 0$, $\dfrac{a^m}{a^n}$ equals:

According to Formula 4, $(a^m)^n$ equals:

According to Formula 5, $(ab)^n$ equals:

For $a \neq 0$, $a^0$ equals:

For $a \neq 0$ and $n \in N$, $a^{-n}$ equals:

Value of $2^5 \cdot 2^6$ is:

Value of $\dfrac{3^5}{3^3}$ is:

Value of $\dfrac{3^3}{3^5}$ is:

$\left(\dfrac{3}{4}\right)^3$ equals:

$(4^2)^7$ equals:

$(a^2 b^3)^5$ equals:

Value of $6^0$ is:

Value of $(-6)^0$ is:

Value of $4^{-1}$ is:

Value of $\left(\dfrac{1}{3}\right)^{-1}$ is:

Value of $\dfrac{a^7 \cdot a^3}{a^4}$ is:

Consider the following statements:

If $x^n = a$ where $n \in N,\, n > 1$, then $x$ is called:

Fourth roots of $16$ are:

Cube root of $-27$ is:

The $n$-th root of $0$ is:

Does $-9$ have a square root in $R$?

For $a > 0$, the principal $n$-th root of $a$ is denoted $\sqrt[n]{a}$ and is:

For $a < 0$ and $n \in N,\, n > 1$ odd, $\sqrt[n]{a}$ equals:

For $a < 0$ and $n$ even, the $n$-th root of $a$:

Value of $-\sqrt[3]{27}$ is:

According to Formula 8, $(\sqrt[n]{a})^m$ equals (for $a > 0$):

Principal value of $\sqrt{4}$ is:

Value of $\sqrt[3]{-8}$ is:

For $a > 0$ and $n \in N,\, n > 1$, $a^{1/n}$ is defined as:

For $a > 0$, $a^{m/n}$ equals:

Value of $8^{2/3}$ is:

Value of $27^{2/3}$ is:

Value of $16^{3/4}$ is:

For $a > 0,\, b > 0$ and $r,s \in Q$, $a^r \cdot a^s$ equals:

For $a > 0$ and $r,s \in Q$, $(a^r)^s$ equals:

For $a, b > 0$ and $r \in Q$, $\left(\dfrac{a}{b}\right)^r$ equals:

If $a^x = 1$ where $a > 0$ and $a \neq 1$, then:

If $a^x = a^y$ where $a > 0$ and $a \neq 1$, then:

If $a^x = b^x$ where $\dfrac{a}{b} > 0$ and $x \neq 0$, then:

If $a^x = b,\, b^y = c,\, c^z = a$, then $xyz$ equals:

If $x^{x\sqrt{x}} = (x\sqrt{x})^x$, then $x$ equals:

If $a^x = b^y = c^z$ and $b^2 = ac$, then $\dfrac{1}{x} + \dfrac{1}{z}$ equals:

If $a^{1/x} = b^{1/y} = c^{1/z}$ and $abc = 1$, then $x + y + z$ equals:

Solving $4^x - 3 \cdot 2^{x+2} + 32 = 0$, the values of $x$ are:

The value of $\left(\dfrac{b}{x^a}\right)^{b+c} \cdot \left(\dfrac{c}{x^b}\right)^{c+a} \cdot \left(\dfrac{a}{x^c}\right)^{a+b}$-style identity in Example 13 simplifies to:

If $a = 2 + 2^{1/3} + 2^{2/3}$, then $a^3 - 6a^2 + 6a - 2$ equals:

Simplified value of $\dfrac{1}{1 + a^{y-x} + a^{z-x}} + \dfrac{1}{1 + a^{x-y} + a^{z-y}} + \dfrac{1}{1 + a^{x-z} + a^{y-z}}$ is:

Consider the following:

If $9^x = 27^y$, then the relation between $x$ and $y$ is:

Solving $3^{2x+2} + 27^{x+1} = 36$, value of $x$ is:

Value of $(a^{1/3} - b^{1/3})(a^{2/3} + a^{1/3}b^{1/3} + b^{2/3})$ is:

If $a^p = b,\, b^q = c$ and $c^r = a$, then $pqr$ equals:

If $a^x = b,\, b^y = c$ and $c^z = 1$ (with $a > 0,\, a \neq 1$), then $xyz$ equals:

If $a^x = y^b = z^c$ and $xyz = 1$, then $ab + bc + ca$ equals (when $a, b, c$ chosen so that $a^x \cdot b^y \cdot c^z = 1$ identity holds):

Value of $\sqrt[3]{(a^3)^2 \cdot \sqrt[3]{a^6}}$-type simplification reduces to:

If $a^m \cdot a^n = (a^m)^n$, then $m(n-2) + n(m-2)$ equals:

If $\sqrt[x]{a} = \sqrt[y]{b} = \sqrt[z]{c}$ and $abc = 1$, then $x + y + z$ equals:

If $b = 1 + 3^{1/3} + 3^{2/3}$, then $b^3 - 3b^2 - 6b - 4$ equals:

If $a + b + c = 0$, then $\dfrac{1}{x^a + x^{-b} + 1} + \dfrac{1}{x^b + x^{-c} + 1} + \dfrac{1}{x^c + x^{-a} + 1}$ equals:

If $a^p = b^q$ and $b = a^2$, then the relation between $p$ and $q$ is:

Logarithm originated from the two Greek words:

The Greek word "Logos" means:

The Greek word "arithmos" means:

If $a^x = b$ where $a > 0$ and $a \neq 1$, then $x$ equals:

If $x = \log_a b$, then $b$ equals:

Conditions on the base of $\log_a b$ are:

The argument $b$ of $\log_a b$ must satisfy:

Value of $\log_2 64$ is:

Value of $\log_8 64$ is:

Value of $\log_a a$ is:

Value of $\log_a 1$ is:

$\log_a (M \cdot N)$ equals:

$\log_a \left(\dfrac{M}{N}\right)$ equals:

$\log_a (M^N)$ equals:

The base-change formula states $\log_a M$ equals: