Exponents are one of the fundamental operations in algebra. On the YÖS exam, problems with exponents appear both in simple form and as part of more complex expressions. Understanding the properties of exponents is essential for working with exponential equations and logarithms.
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Definition: A power of a number $a$ with natural exponent $n$ is the product of $n$ identical factors equal to $a$.
$$a^n = \underbrace{a \cdot a \cdot a \cdots a}_{n \text{ раз}}$$
where $a$ is the base of the power and $n$ is the exponent.
Examples:- $2^3 = 2 \cdot 2 \cdot 2 = 8$- $5^2 = 5 \cdot 5 = 25$- $(-2)^4 = (-2) \cdot (-2) \cdot (-2) \cdot (-2) = 16$- $(-2)^3 = (-2) \cdot (-2) \cdot (-2) = -8$
Important remarks:- $a^1 = a$ for any $a$- If the base is negative and the exponent is even, the result is positive- If the base is negative and the exponent is odd, the result is negative
Property: When multiplying powers with the same base, the exponents are added.
$$a^m \cdot a^n = a^{m+n}$$
Examples:$$2^3 \cdot 2^5 = 2^{3+5} = 2^8 = 256$$$$x^2 \cdot x^4 = x^{2+4} = x^6$$$$3^{-2} \cdot 3^5 = 3^{-2+5} = 3^3 = 27$$
Proof:$$a^m \cdot a^n = \underbrace{a \cdot a \cdots a}_{m \text{ раз}} \cdot \underbrace{a \cdot a \cdots a}_{n \text{ раз}} = \underbrace{a \cdot a \cdots a}_{m+n \text{ раз}} = a^{m+n}$$
Property: When dividing powers with the same base, the exponents are subtracted.
$$\frac{a^m}{a^n} = a^{m-n} \quad (a \neq 0)$$
Examples:$$\frac{2^7}{2^3} = 2^{7-3} = 2^4 = 16$$$$\frac{x^8}{x^2} = x^{8-2} = x^6$$$$\frac{5^{-2}}{5^{-5}} = 5^{-2-(-5)} = 5^3 = 125$$
Property: When raising a power to a power, the exponents are multiplied.
$$(a^m)^n = a^{m \cdot n}$$
Examples:$$(2^3)^2 = 2^{3 \cdot 2} = 2^6 = 64$$$$(x^2)^5 = x^{2 \cdot 5} = x^{10}$$$$((3^2)^3)^2 = 3^{2 \cdot 3 \cdot 2} = 3^{12}$$
Property: The power of a product is equal to the product of the powers.
$$(ab)^n = a^n \cdot b^n$$
Examples:$$(2 \cdot 3)^3 = 2^3 \cdot 3^3 = 8 \cdot 27 = 216$$$$(xy)^4 = x^4 y^4$$$$(2x)^3 = 2^3 \cdot x^3 = 8x^3$$
Property: The power of a quotient is equal to the quotient of the powers.
$$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \quad (b \neq 0)$$
Examples:$$\left(\frac{2}{3}\right)^2 = \frac{2^2}{3^2} = \frac{4}{9}$$$$\left(\frac{x}{y}\right)^3 = \frac{x^3}{y^3}$$$$\left(\frac{5}{2}\right)^{-1} = \frac{2}{5}$$
Definition: For any nonzero number $a$:
$$a^0 = 1$$
Justification: Using the property $\frac{a^m}{a^m} = a^{m-m} = a^0$:$$\frac{a^m}{a^m} = 1 \quad \text{и} \quad \frac{a^m}{a^m} = a^{m-m} = a^0$$
Therefore, $a^0 = 1$
Examples:- $5^0 = 1$- $(-7)^0 = 1$- $(3x)^0 = 1$ (for $x \neq 0$)- $x^0 = 1$ (for $x \neq 0$)
Important: $0^0$ is undefined!
Definition: For any nonzero number $a$ and natural number $n$:
$$a^{-n} = \frac{1}{a^n}$$
Justification: Using the property $\frac{a^m}{a^n} = a^{m-n}$:$$a^0 = a^{n-n} = \frac{a^n}{a^n} = a^{-n} \cdot a^n$$
Therefore:$$a^{-n} = \frac{1}{a^n}$$
$$2^{-3} = \frac{1}{2^3} = \frac{1}{8}$$$$5^{-2} = \frac{1}{5^2} = \frac{1}{25}$$$$x^{-4} = \frac{1}{x^4}$$$$\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$$
$$a^{-m} \cdot a^n = a^{-m+n} = a^{n-m}$$$$\frac{a^{-m}}{a^n} = a^{-m-n}$$$$(a^{-m})^n = a^{-mn}$$
Example:$$\frac{2^{-3} \cdot 2^5}{2^{-2}} = 2^{-3+5-(-2)} = 2^{-3+5+2} = 2^4 = 16$$
Definition: For natural $n > 1$:
$$a^{1/n} = \sqrt[n]{a}$$
(the $n$-th root of $a$)
Examples:- $8^{1/3} = \sqrt[3]{8} = 2$ (since $2^3 = 8$)- $16^{1/2} = \sqrt{16} = 4$ (since $4^2 = 16$)- $32^{1/5} = \sqrt[5]{32} = 2$ (since $2^5 = 32$)
Definition:
$$a^{m/n} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m$$
(both forms are equivalent)
Examples:$$8^{2/3} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4$$
or
$$8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$$
$$27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$$$$16^{3/2} = \left(\sqrt{16}\right)^3 = 4^3 = 64$$
- All properties of powers hold for fractional exponents as well- For even $n$ and a negative base, the expression is undefined in the real numbers: $$(-8)^{1/2} \text{ не определено в } \mathbb{R}$$
- For odd $n$, a fractional power is also defined for negative bases: $$(-8)^{1/3} = \sqrt[3]{-8} = -2$$
If an equation contains powers with the same base, we use the property: if $a^m = a^n$ and $a > 0, a \neq 1$, then $m = n$.
Example 1: Solve $2^{x+1} = 16$
$$2^{x+1} = 2^4$$$$x + 1 = 4$$$$x = 3$$
Example 2: Solve $3^{2x} = 27$
$$3^{2x} = 3^3$$$$2x = 3$$$$x = 1.5$$
Example 3: Solve $\left(\frac{1}{5}\right)^{x} = 125$
$$5^{-x} = 5^3$$$$-x = 3$$$$x = -3$$
Example 4: Solve $4^x = 8$
Note that $4 = 2^2$ and $8 = 2^3$:$$(2^2)^x = 2^3$$$$2^{2x} = 2^3$$$$2x = 3$$$$x = 1.5$$
Example 5: Solve $9^x \cdot 27 = 3^5$
$$9 = 3^2, \quad 27 = 3^3$$$$(3^2)^x \cdot 3^3 = 3^5$$$$3^{2x} \cdot 3^3 = 3^5$$$$3^{2x+3} = 3^5$$$$2x + 3 = 5$$$$2x = 2$$$$x = 1$$
Example 6: Solve $4^x - 3 \cdot 2^x + 2 = 0$
Note that $4^x = (2^2)^x = 2^{2x} = (2^x)^2$
Let $t = 2^x$, then:$$t^2 - 3t + 2 = 0$$$$(t - 1)(t - 2) = 0$$$$t = 1 \text{ или } t = 2$$
Case 1: $2^x = 1 = 2^0 \Rightarrow x = 0$
Case 2: $2^x = 2 = 2^1 \Rightarrow x = 1$
Answer: $x = 0$ or $x = 1$
The function $f(x) = a^x$ is:- increasing when $a > 1$- decreasing when $0 < a < 1$
Example 1: Compare $2^5$ and $2^7$
Since $2 > 1$ and the function $2^x$ is increasing, from $5 < 7$ we get $2^5 < 2^7$
Example 2: Compare $\left(\frac{1}{2}\right)^3$ and $\left(\frac{1}{2}\right)^5$
Since $0 < \frac{1}{2} < 1$ and the function $\left(\frac{1}{2}\right)^x$ is decreasing, from $3 < 5$ we get $\left(\frac{1}{2}\right)^3 > \left(\frac{1}{2}\right)^5$
Example 3: Compare $2^{10}$ and $4^4$
$$4^4 = (2^2)^4 = 2^8$$
Since $2^{10} = 2^8 \cdot 2^2 = 2^8 \cdot 4$ and $2^8 < 2^{10}$, we have $4^4 < 2^{10}$
or: $2^{10}$ and $2^8$, where $10 > 8$, hence $2^{10} > 2^8 = 4^4$
Example 4: Compare $\left(\frac{3}{5}\right)^{100}$ with 1
Since $0 < \frac{3}{5} < 1$, we have $\left(\frac{3}{5}\right)^{100} < 1$
Example 5: Compare $2^{-5}$ with $\left(\frac{1}{2}\right)^{5}$
$$2^{-5} = \frac{1}{2^5} = \left(\frac{1}{2}\right)^5$$
They are equal!
Statement: Compute $(2^3)^2 \cdot 2^{-4} : 2^2$
Solution:
$$(2^3)^2 \cdot 2^{-4} : 2^2 = 2^{3 \cdot 2} \cdot 2^{-4} : 2^2$$$$= 2^6 \cdot 2^{-4} : 2^2$$$$= 2^{6-4} : 2^2$$$$= 2^2 : 2^2$$$$= 2^{2-2} = 2^0 = 1$$
Answer: 1
Statement: Compute $16^{3/4}$
$$16^{3/4} = (16^{1/4})^3 = (\sqrt[4]{16})^3 = 2^3 = 8$$
Or:$$16^{3/4} = \sqrt[4]{16^3} = \sqrt[4]{4096} = 8$$
Answer: 8
Statement: Solve $5^{2x-1} = 25$
$$5^{2x-1} = 25 = 5^2$$$$2x - 1 = 2$$$$2x = 3$$$$x = 1.5$$
Answer: $x = 1.5$
Statement: Solve $9^x + 3^{x+1} = 12$
$$9^x = (3^2)^x = 3^{2x} = (3^x)^2$$$$3^{x+1} = 3 \cdot 3^x$$
Let $t = 3^x$, where $t > 0$:$$t^2 + 3t = 12$$$$t^2 + 3t - 12 = 0$$$$(t + 4)(t - 3) = 0$$
$t = -4$ is not valid (since $t > 0$)
$t = 3$:$$3^x = 3 = 3^1$$$$x = 1$$
Answer: $x = 1$
Statement: Arrange in increasing order: $2^5, 4^2, 8^{4/3}$
Let us reduce all of them to the same base (2):- $2^5 = 2^5 = 32$- $4^2 = (2^2)^2 = 2^4 = 16$- $8^{4/3} = (2^3)^{4/3} = 2^{3 \cdot 4/3} = 2^4 = 16$
Note that $4^2 = 8^{4/3} = 2^4 = 16$
In increasing order: $4^2, 8^{4/3}, 2^5$ or $2^4, 2^4, 2^5$
Answer: $4^2 = 8^{4/3} < 2^5$
Wrong: $2^3 \cdot 2^4 = 2^{3 \cdot 4} = 2^{12}$
Correct: $2^3 \cdot 2^4 = 2^{3+4} = 2^7$
Wrong: $(a + b)^2 = a^2 + b^2$
Correct: $(a + b)^2 = a^2 + 2ab + b^2$
Wrong: $-2^2 = 4$ (this is a mistake!)
Correct: $-2^2 = -(2^2) = -4$, while $(-2)^2 = 4$
Wrong: $8^{2/3} = 8^2 : 3 = 64 : 3 \approx 21$
Correct: $8^{2/3} = (8^{1/3})^2 = 2^2 = 4$
When solving $2^x = -4$, some mistakenly say there are no solutions only because they forgot about the property of the exponential function (it is always positive).
Correct: $2^x > 0$ for all $x$, therefore the equation $2^x = -4$ has no solutions.
Wrong: In the equation $4^x - 2 \cdot 2^x + 1 = 0$, substituting $t = 2^x$ and forgetting that $4^x = t^2$ rather than $t$
Correct: $4^x = (2^2)^x = 2^{2x} = (2^x)^2 = t^2$
### Tip 1: Memorize powers of 2, 3, 5This will speed up your calculations:- $2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64, 2^7=128, 2^8=256, 2^9=512, 2^{10}=1024$- $3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243$- $5^1=5, 5^2=25, 5^3=125, 5^4=625$
### Tip 2: When solving exponential equations, look for a common baseThis is the most universal method for simple equations.
### Tip 3: Use variable substitution for complex equationsIf an equation contains $a^x$ and $a^{2x}$ (or something similar), variable substitution often simplifies the solution.
### Tip 4: Always check the restrictions on the exponentRemember that negative and fractional exponents impose restrictions on the base.
### Tip 5: When comparing powers, use the monotonicity of the functionIf the base is the same, a larger exponent yields a larger result only if the base is > 1.
### Tip 6: Be careful with a negative base$(-2)^3 \neq -(2^3)$, but $(-2)^2 = 2^2$
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