Exponents and Logarithms

Graphing Exponential and Logarithmic Functions

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Exponents and logarithms have numerous applications in various fields, including mathematics, science, engineering, finance, and computer science. Here are some examples:

Compound interest: Compound interest is a type of exponential growth that involves the use of exponents. It is used to calculate the interest earned on a principal amount over a specific period of time.
Population growth: The growth of a population can also be modeled using exponents. The exponential growth model is used to predict the future population size based on the current growth rate.
Electrical engineering: In electrical engineering, exponents are used to describe the relationship between voltage, current, and resistance in Ohm's Law.
Chemistry: The pH scale used in chemistry is based on logarithms. The pH of a substance is the negative logarithm of the concentration of hydrogen ions in the solution.
Data compression: In computer science, logarithms are used in data compression algorithms to compress large amounts of data into smaller files.
Signal processing: Exponential functions and logarithmic functions are used in signal processing to describe the decay rate of signals over time.
Music: In music, logarithmic functions are used to measure the intensity of sound waves, which is expressed in decibels (dB).
Probability: In probability theory, logarithmic functions are used to express the odds of an event occurring.

These are just a few examples of how exponents and logarithms are used in various fields.

Solved Problems On Exponents (With explanations):

Problem 1: Simplify the expression: $3^4\ *\ 3^2$

Solution:

When we multiply two powers with the same base, we add their exponents. Therefore:

$3^4\ *\ 3^2=3^{4+2}=3^{6}=729$

Therefore, the simplified expression is 729.

Problem 2: Solve for x: $4^x = 64$

Solution:

To solve for x, we can take the logarithm of both sides of the equation. Since the base of the exponent is 4, we can use the logarithm base 4:

$log4(4^x) = log4(64)$

Using the logarithmic identity $logb(b^x) = x$ , we can simplify the left side of the equation:

$x = log4(64) = log(64) / log(4)$

Using a calculator, we can evaluate the logarithms:

$x = 3$

Therefore, the solution to the equation is x = 3.

Problem 3: Simplify the expression: $log4(16) - log4(2)$

Solution:

Using the logarithmic identity $logb(x/y) = logb(x) - logb(y)$ , we can simplify the expression:

$log4(16) - log4(2) = log4(16/2) = log4(8)$

Since $4^2 = 16$ and $4^{\frac{1}{2}} = 2$ , we know that $8 = 4^{\frac{3}{2}}$ .

Therefore:

$log4(8) = log4(4^{\frac{3}{2}}) = \frac{3}{2}$

Therefore, the simplified expression is 3/2.

Problem 4:

$Solve\ for\ x: 2^x = 16$

Solution:

To solve for x, we can rewrite 16 as a power of 2. Since 16 is equal to $2^4$ , we can set the equation as $2^{x} =2^{4}$ .

Using the property of exponentiation, we know that if two powers with the same base are equal, then their exponents must also be equal. Therefore, x = 4.

Hence, the solution to the equation $2^{x} =16$ .

Problem 5:

$Solve\ for\ y: (\frac{1}{3})^y = 9$

Solution:

To solve for y, we need to express 9 as a power of 1/3. We know that 1/3 is equal to $(1/3)^1$ , and 9 is equal to $(\frac{1}{3})^{-2}$ ..

We can rewrite the equation as

$(\frac{1}{3})^y = (\frac{1}{3})^{-2}.$

Again, using the property of exponentiation, we equate the exponents:

y = -2.

Therefore, the solution to the equation $(\frac{1}{3})^y = 9$ is y = -2.

Problem 6:

Solve for z:

$5^z = 125$

Solution:

In this equation, 125 is equal to 5^3. Therefore, we can rewrite the equation as $5^z = 5^3.$

By equating the exponents, we find that z = 3.

Thus, the solution to the equation $5^z = 125 \ \ is \ \ z = 3.$

Problem 7:

Solve for 'a' :

$(2^a)(2^3) = 64$

Solution:

We can simplify the left side of the equation by multiplying the powers of 2:

$(2^a)(2^3) = 2^{(a+3)} = 64.$

Since 64 is equal to $2^{6}$ , we can rewrite the equation as $2^{(a+3)} = 2^6$ .

By equating the exponents, we have a+3 = 6.

Subtracting 3 from both sides of the equation, we get:

$a+3{\color{Red} -3}=6{\color{Red} -3}$

a = 3.

Hence, the solution to the equation

$(2^a)(2^3) = 64 \ is\ a = 3.$

Problem 8:

Solve for x::

$3^{(2x+1)} = 81$

Solution:

To solve for x, we can start by expressing 81 as a power of 3. We know that 81 is equal to $3^4$ .

So, we can rewrite the equation as $3^{(2x+1)} = 3^4.$

Using the property of exponentiation, we equate the exponents:

2x + 1 = 4

Subtracting 1 from both sides, we get:

2x = 3

Dividing both sides by 2, we find

$x = \frac{3}{2}$

Hence, the solution to the equation

$3^{(2x+1)} = 81 \ {\color{Red} is} \ x = \frac{3}{2}.$

Problem 9:

Solve for y:

$(4^y)(4^2y) = 64^3$

Solution:

To solve for y, we can simplify the left side of the equation by multiplying the powers of 4:

$(4^y)(4^2y) = 4^{(y + 2y)} = 4^{(3y)}.$

We can rewrite the equation as

$4^(3y) = (4^3)^3$

Simplifying further, we get

$4^{(3y)} = 64^3$

Since both sides have the same base, we equate the exponents:

3y = 3

Dividing both sides by 3, we find:

y = 1

$Therefore, the \ solution\ to\ the \ equation \ (4^y)(4^2y) = 64^3 \ {\color{Red} is}\ y = 1.$

Problem 10:

Solve for z:

$(9^z)(27^z) = 81^4$

Solution:

To solve for z, we can simplify the left side of the equation by multiplying the powers of 9 and 27:

$(9^z)(27^z) = (3^2z)(3^3z) = 3^{(2z + 3z)} = 3^{(5z)}.$

$We\ can\ rewrite\ the\ equation\ as\ 3^{(5z)} = (3^4)^4.$

$Simplifying \ further,\ we\ get\ 3^{(5z)} = 81^4.$

By equating the exponents, we have:

$5z = 4$

Dividing both sides by 5, we find:

$z = \frac{4}{5}$

$Hence,\ the\ solution\ to\ the\ equation\ (9^z)(27^z) = 81^4 \ {\color{Red} \\ is} \ z = \frac{4}{5}.$

Solved Problems On Exponents:

Solve the following exponential equation for x:

Problem 1:

$\ 2^{3x+1}=64$

Solution:

$\ 2^{3x+1}=64$

$2^{3x+1}=2^{6}$

$3x+1=6$

$3x=6-1$

$3x=5$

$x=\frac{5}{3}$ ,answer

Problem 2:

$3^{2x-4}.9^{x+1}=27$

Solution:

$3^{2x-4}.9^{x+1}=27$

$3^{2x-4}.(3^2)^{x+1}=27$

$3^{2x-4}\ .\ 3^2^{(x+1)}=27$

$3^{2x-4+2(x+1)}=27$

$3^{4x-2}=27$

$3^{4x-2}=3^3$

$4x-2=3$

$4x-2{\color{Red} +2}=3{\color{Red} +2}$

$4x=5$

$x=\frac{5}{4}$ ,answer

Problem 3:

$5^{2x+3}\ .\ 125^{-x}=\frac{1}{25}$

Solution:

$5^{2x+3}\ .\ 125^{-x}=\frac{1}{25}$

$5^{2x+3}\ .\ (5^3)^{-x}=\frac{1}{25}$

$5^{2x+3}\ .\ 5^^{-3x}=\frac{1}{25}$

$5^{2x+3-3x}=\frac{1}{25}$

$5^{-x+3}=\frac{1}{25}$

$5^{-x+3}=\frac{1}{5^2}$

$5^{-x+3}=5^{-2}$

$-x+3=-2$

$- x = - 2 - 3$

$- x = - 5$

$x =$ ,answer

Problem 4:

$2^{(3x - 4)} = 8$

Solution:

$2^{(3x - 4)} = 8$

$2^{(3x - 4) }= 2^3$

$3x - 4 = 3$

$3x - 4 {\color{Red} + 4} = 3 {\color{Red} + 4}$

$3x = 7$

$\frac{3x}{3} = \frac{7}{3}$

$x = \frac{7}{3}$ ,answer

Problem 5:

$3^{(2x + 1)} = 27$

Solution:

$3^{(2x + 1)} = 27$

$3^{(2x + 1)} = 3^3$

$2x + 1 = 3$

$2x + 1 {\color{Red} - 1} = 3 {\color{Red} - 1}$

$2x = 2$

$\frac{2x}{2} = \frac{2}{2}$

$x = 1$ ,answer

Problem 6:

$5^{(x + 2)} = 125$

Solution:

$5^{(x + 2)} = 125$

$5^{(x + 2)} = 5^3$

Setting the exponents equal to each other:

$x + 2 = 3$

$x + 2 {\color{Red} - 2} = 3 {\color{Red} - 2}$

$x = 1$ ,answer

Problem 7:

$10^{(x - 1)} = \frac{1}{100}$

Solution:

$10^{(x - 1)} = \frac{1}{100}$

$10^{(x - 1)} = 10^{(-2)}$

$x - 1 = -2$

$x - 1 {\color{Red} + 1} = -2 {\color{Red} + 1}$

$x = -1$ ,answer

Problem 7:

$2^x =\frac{ 1}{8}$

Solution:

$2^x =\frac{ 1}{8}$

$2^x = 2^{^(-3)}$

$x = -3$ ,answer

Problem 8:

$3(2x+2)-2(2x+1)=0$

Solution:

$2^{x+1}(3*2-2)=0$

$2^{x+1}(6-2)=0$

$2^{x+1}*4=0$ ,answer

Since is nonzero, for the equation to be true, ${2^{x+1}$ must be equal to . However, there is no value of that will make $2^{x + 1}$ equal to .

Hence, there is no solution to the equation.

Problem 9:

$4^{3x} . \ 4^{2x+1}$

Solution:

$4^{3x} .\ 4^{2x+1}=4^{3x+2x+1}$

$4^{3x} .\ 4^{2x+1}=4^{3x+2x+1} =4^{5x+1}$ ,answer

Problem 10:

$(\frac{2^{4}}{2^{2}})^{3}$

Solution:

$(\frac{2^{4}}{2^{2}})^{3}=(\frac{2^{4-2}}{1})^{3}$

$(\frac{2^{4}}{2^{2}})^{3}=(\frac{2^{4-2}}{1})^{3}=(\frac{2^{2}}{1})^{3}$

$(\frac{2^{4}}{2^{2}})^{3}=(\frac{2^{4-2}}{1})^{3}=(\frac{2^{2}}{1})^{3}=(2^{2})^{3}$

$(\frac{2^{4}}{2^{2}})^{3}=(\frac{2^{4-2}}{1})^{3}=(\frac{2^{2}}{1})^{3}=(2^{2})^{3}=2^{6}$

$(\frac{2^{4}}{2^{2}})^{3}=(\frac{2^{4-2}}{1})^{3}=(\frac{2^{2}}{1})^{3}=(2^{2})^{3}=2^{6}=64$ ,answer

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