Unit 4 Exponential & Logarithmic Practice Quiz

Exponential functions are characterized by a constant raised to a variable power. The expression y = ab^x correctly reflects this property, where a is the initial value and b is the base.

Substituting x = 2 into the function f(x) = 3^x gives f(2) = 3^2, which equals 9. This demonstrates how to evaluate simple exponential expressions.

Exponential growth functions increase rapidly as x increases, which is best illustrated by a graph that rises steeply to the right. The other options describe linear, quadratic, or constant behaviors.

For exponential functions of the form y = ab^x, as x becomes very negative, the function approaches 0, hence the horizontal asymptote is y = 0. This property distinguishes it from other types of functions.

The inverse of an exponential function is a logarithmic function with the same base. Since f(x) = 2^x, its inverse is given by fā»Ā¹(x) = logā‚‚(x).

Express 16 as a power of 2; 16 is 2^4. Therefore, 2^(x+1) = 2^4 implies that x + 1 = 4, so x = 3. This demonstrates the method of equating exponents when the bases are the same.

For logarithmic functions, the argument must be positive. Since the argument here is (x - 3), we require x - 3 > 0, hence x > 3. This is a key property of logarithms.

The product rule for logarithms states that logā‚(x) + logā‚(y) = logā‚(xy). This property is fundamental when simplifying or combining logarithmic expressions.

By rewriting the logarithmic equation in exponential form, we have x = 3ā“, which equals 81. This illustrates the conversion between logarithmic and exponential forms.

The inverse of an exponential function is found by taking the logarithm with the same base. Hence, the inverse of f(x) = 5^x is fā»Ā¹(x) = logā‚…(x).

Exponential growth functions have a base greater than 1, resulting in increasing values as x increases. Conversely, exponential decay functions use a base between 0 and 1, causing the function to decrease.

Using the exponent rule e^(2x) = (eĀ²)^x simplifies the expression. This property of exponents allows us to view exponential functions in alternative forms.

By using logarithm properties, ln(1/e^x) can be rewritten as ln(e^(-x)), which simplifies directly to -x. This demonstrates the power rule for logarithms in action.

Since logā‚‚(8) equals 3 and logā‚‚(4) equals 2, their sum is 3 + 2 = 5. This problem reinforces the evaluation of logarithms with known powers of the base.

Substituting x = 2 into the function yields 3^(2 - 2) = 3^0, which is equal to 1. This demonstrates the critical principle that any non-zero number raised to the zero power is 1.

Rewrite 4 as 2Ā² so that 4^(x + 1) becomes (2Ā²)^(x + 1) = 2^(2x + 2). Since 32 is 2^5, equate the exponents: 2x + 2 = 5, which gives x = 3/2. This method relies on expressing both sides with a common base.

Recognize that 8 can be written as 2Ā³. Thus, g(x) = 8^x becomes (2Ā³)^x = 2^(3x), which is exactly f(x). This demonstrates the importance of rewriting expressions to reveal underlying equivalences.

Express 27 as 3Ā³. Setting 3^(2x) equal to 3Ā³ allows you to equate the exponents: 2x = 3, hence x = 3/2. This problem reinforces solving exponential equations by comparing powers.

The power rule of logarithms states that log_b(x^c) equals c log_b(x). Here, log_b(xĀ²) becomes 2 log_b(x), which is a direct application of the power rule.

Since logā‚‚(8) = 3 and 8 = 2Ā³, for 16 = 2ā“ the logarithm f(16) is logā‚‚(16) which equals 4. This problem shows how logarithms translate exponents into numerical values.