Evaluating Logarithms

Evaluating Logarithms

• Logarithms determine the power to which a base must be raised to obtain a specific number.
• To evaluate log base 2 of 4, determine what power of 2 equals 4; since 2^2 = 4, log base 2 of 4 is 2.
• Log base 3 of 9 equals 2 because 3 squared is 9.
• Log base 4 of 16 is 2 because 4 squared is 16.
• Log base 3 of 27 is 3 because 3 cubed is 27.
• Log base 2 of 32 is 5 because 2 to the fifth power is 32.
• Log base 5 of 125 is 3 because 5 cubed is 125.
• Log base 6 of 36 is 2 because 6 squared is 36.
• Log base 2 of 64 is 6 because 2 to the sixth power is 64.
• Log base 3 of 81 is 4 because 3 to the fourth power is 81.
• Log base 7 of 49 is 2, based on 7 squared equaling 49.
• Log base 3 of 1 is always 0, since any number raised to the 0 power equals 1.
• When the base isn't explicitly stated in the log, it is assumed to be base 10.
• Log of 10 is 1 because 10 to the first power is 10.
• Log of 100 is 2 because 10 squared is 100.
• Log of 1,000 is 3 because 10 cubed is 1,000.
• Log of 1 million is 6 because 10 to the sixth power is 1 million.
• Log of 0.1 is -1.
• Log of 0.01 is -2.
• Log of 0.001 is -3.
• Log of 0.00001 is -5.
• The log of 0 and the log of a negative number are undefined; the value inside a log must be greater than 0.
• For log base 4 of 16, the answer is 2 since 4 squared equals 16.
• Log base 4 of 1/16 is -2, because 4 to the power of -2 is 1/16.
• Log base 16 of 4 is 1/2.
• Log base 16 of 1/4 is -1/2.
• If a log contains a fraction, the answer will be negative.
• If the number inside the log is larger than the base, the answer is typically a number greater than 1.
• If the base is larger than the number inside the log, expect a fraction (number less than 1).
• Log base 6 of 216 is 3, since 6 cubed is 216.
• Log base 6 of 1/216 is -3.
• Log base 216 of 6 is 1/3.
• Log base 216 of 1/6 is -1/3.

Change of Base Formula

• The change of base formula allows you to convert a logarithm from one base to another: log base a of B = log B / log A.
• The new base can be any number.
• For example, log base 4 of 16, which equals 2, can be calculated using a calculator by inputting log 16 / log 4.
• To convert log base 5 of (x - y) into a natural log, apply the change of base formula to get log(x-y)/log(5), and then convert to natural log: ln(x-y)/ln(5).

Properties of Logs

• Combining logs with addition: log A + log B = log (A * B)
• Combining logs with subtraction: log A - log B = log (A / B)
• Power rule: log A^2 = 2 log A

Condensing Logarithmic Expressions

• For log X + log Y - log Z, the condensed form is log (XY/Z): positive signs indicate multiplication in the numerator, and negative signs indicate division in the denominator.
• For log X - log Y + log Z - log R, the condensed form is log (XZ/YR).
• To condense 2 log X + 3 log Y - 4 log Z, first move the coefficients to become exponents: log(X^2) + log(Y^3) - log(Z^4), then combine: log (X^2 * Y^3 / Z^4).
• For 1/2 log X - 1/3 log Y + 1/4 log Z, first move the coefficients: log(X^(1/2)) - log(Y^(1/3)) + log(Z^(1/4)), condense to get log (X^(1/2) * Z^(1/4) / Y^(1/3)), and convert to radical form: log (√X * ⁴√Z / ³√Y).

Expanding Logarithmic Expressions

• For log (R^2 * S^5 / Z^6), expand to: log(R^2) + log(S^5) - log(Z^6), then move exponents to the front: 2 log R + 5 log S - 6 log Z.
• For log (³√(X^2 * Y / Z^4)), rewrite as log ((X^2 * Y / Z^4)^(1/3)), move the exponent to the front: 1/3 [log(X^2) + log(Y) - log(Z^4)], distribute: (1/3)log(X^2) + (1/3)log(Y) - (4/3)log(Z), distribute again: 2/3 log X + 1/3 log Y - 4/3 log Z.
• To expand log (X^2 * √Y / ³√Z^4), rewrite as log (X^2 * Y^(1/2) / Z^(4/3)), then expand to log(X^2) + log(Y^(1/2)) - log(Z^(4/3)), and move exponents: 2 log X + 1/2 log Y - 4/3 log Z.

Simplifying Natural Log Expressions

• Natural log of 1 is always 0, regardless of the base.
• Natural log of e is always equal to 1.
• Ln(e^5) = 5 ln(e) = 5 * 1 = 5, where the exponent can be moved to the front.
• e^(ln 7) simplifies to 7, because the base e and natural log base e cancel each other out.
• e^(3 ln x) is equivalent to e^(ln x^3), which simplifies to x^3 by the same base-cancellation principle.
• 2 * e^(ln y^4) simplifies to 2y^4 (moving the three makes e to the ln cancel).
• 5^(log base 5 of y^8) simplifies to y^8.
• 3^(log base 3 of (x^2 + 6)) simplifies to x^2 + 6.

Simplifying Log Expressions

• log base 3 of 6 - log base 3 of 2 can be combined into log base 3 of (6/2) which simplifies to log base 3 of 3, which equals 1.
• log base 2 of 12 + log base 2 of 24 - log base 2 of 9 condenses to log base 2 of (12*24/9) = log base 2 of 32. Since 2^5 = 32, the expression equals 5.
• log base 7 of 1/6 + log base 7 of 6/49 condenses to log base 7 of (1/6 * 6/49) = log base 7 of 1/49. Since 7^(-2) = 1/49, the answer is -2.

Converting Between Logarithmic and Exponential Forms

• 2^3 = 8: Log base 2 of 8 is equal to 3. The base of the exponent is equal to the base of the log.
• 3^2 = 9: Log base 3 of 9 is equal to 2.
• 5^2 = 25: Log base 5 of 25 is equal to 2.
• a^c = b: Log base a of b is equal to c.
• 2^y = x: Log base 2 of x is equal to y.
• Convert 3^3 = 27 to logarithmic form: log base 3 of 27 = 3.
• Convert 2^4 = 16 to logarithmic form: log base 2 of 16 = 4.
• 3^x = 18 is equivalent to log base 3 of 18 = x.

Solving Logarithmic Equations

• If log base x of 27 = 3, convert to exponential form: x^3 = 27, solve for x by finding the cube root of 27, which is 3.
• If log base 2 of y = 5, convert to exponential form: 2^5 = y, so y = 32.

Solving Logarithmic Equations with Expressions

• If log base 2 of (x + 5) = 3, convert to exponential form: 2^3 = x + 5, so 8 = x + 5, subtract 5 from both sides: x = 3.
• If log base 3 of (2x - 3) = 2, convert to exponential form: 3^2 = 2x - 3, so 9 = 2x - 3, add 3: 12 = 2x, divide by 2: x = 6.
• To solve log base 2 (x-3) + log base 2 (x-1) = 3, first combine the logs into log base 2 ((x-3)(x-1)) = 3, convert to exponential: 2^3 = (x-3)(x-1). Simplify: 8 = x^2 - 4x + 3, set to 0: x^2 - 4x - 5 = 0. Factor: (x-5)(x+1) = 0, therefore x = 5 or x = -1. Because there can be no negatives in a log, 5 is kept and -1 is an extraneous solution (disregarded).
• For log base 3 (2x+1) + log base 3 (x+8) = 3, condense to log base 3 ((2x+1)(x+8)) = 3, rewrite in exponential form: 3^3 = (2x+1)(x+8), simplify: 27 = 2x^2 + 17x + 8, move all to one side: 0 = 2x^2 + 17x - 19. Factor the trinomial: (2x+19)(x-1)=0 . This provides the answers x=1, and x= -19/2 . After checking for extraneous solutions, it it determined that 1 is kept and -19/2 is extraneous and disregarded.
• For log base 2 (x+6) - log base 2 (x-8) = 3, condense to log base 2 ((x+6)/(x-8)) = 3, rewrite in exponential form: 2^3 = (x+6)/(x-8), simplify to 8 = (x+6)/(x-8) or 8x-64=x+6. Continue by isolating x to get an answer of x=10.
• If log base 2 (x+5) = log base 2 (3x-9), since the bases are the same, then x+5=3x-9; isolate x for an answer of x=7.
• Ln x = 5, convert to exponential form e^5 = x.
• Ln x + 3 Ln x = 0, or 4 Ln x = 0, simplifies to Ln x = 0, exponentiate: e^0 = x, therefore x = 1.
• If Ln (x-3) = 2, rewrite e^2 = x-3 and isolate x, for an answer of x = e^2 +3

Solving Exponential Equations

• To solve 8^(x+4) = 16^(2x-3), express both sides with a common base: (2^3)^(x+4) = (2^4)^(2x-3), simplify to 2^(3x+12) = 2^(8x-12), thus 3x+12 = 8x-12, simplify to 5x=24, finally x = 24/5.
• Solve 9^(2x-1) = 27^(3x+6) by expressing both sides in base 3: (3^2)^(2x-1) = (3^3)^(3x+6), simplify: 3^(4x-2) = 3^(9x+18), thus the equation becomes 4x-2 = 9x+18 . Simplify: -5x=20 thus giving a final result of x=-4.
• Solve 2^x=7 by taking the natural log of both sides. This changes the equation into x*Ln2=Ln7. Isolate x by dividing both sides of the equation by Ln2, to get x=Ln7/Ln2. Simplify with a calculator to x=2.80735 approxiamtely. This number can be checked by testing 2^(2.80735) resulting in approximately 7, thus verifying the answer.
• To solve e^(x+2) = 8, take natural log of both sides: Ln(e^(x+2)) = Ln 8, simplify (x+2)Lne = Ln 8. Continue by isolating x: x = Ln 8 -2.
• To solve e^(x+2) = 8, take natural log of both sides: Ln(e^(x+2)) = Ln 8, simplify (x+2)Lne = Ln 8. Continue by isolating x: x = Ln 8 -2.

Finding the Domain of Logarithmic Functions

• For f(x) = Ln(2x-6), the domain requires 2x-6 >0, so solve 2x>6, and divide, thus the domain is x>3. In notation, the domain is (3, infinity).
• To find the domain for Ln (x^2 +2x -15), requires x^2+2x-15>0, and factoring (x+5)(x-3)>0. The critical numbers appear to be -5 and 3. Plot critical numbers of -5 and 3 on a number line. Choosing numbers left of -5, between -5 and 3, and to the right of 3, can determine the equation's nature in those areas. The area should be positive because that is what the equation asked for. Testing those numbers can show those areas: Left of -5 positive, -5 and 3 negative, and right of 3, positive. Since the equation needs to be greater than 0 , but not including 0, a parentheses needs to be used. The final answer should be x= ( negative infinity, -5) U ( 3, positive infinity).

Finding Inverse Functions

• For f(x)= Ln( x+3) -1, solve the inverse function by replacing f(x) with y. Thus gives y= Ln(x+3) -1. To find the inverse switch x and y. Isolate y: e^(x+1)= y+3. Isolating y gives an updated equation of y= e^(x+1) -3. Using inverse notation gives the fianl answer, f^-1(x) = e^(x+1)-3.

Graphing Exponential Functions

• Graph y= 2^(x-3) +1, determine what x-3 should be equal to. To do that set x-3= 0 and 1. The answer is x= 3 and 4. Create a table, and using these answers graph, find the horizontal asymptote. The location of the horizontal asymtote is always the value "added", if present, in the original function's structure.
• Start with e^(x+1) -2 and create a graph: To do that, determine what x+1 should be equal to.To do that set x+1= 0 and 1, the result being x=-1 and x=0. Use these values to plot points into table. Also locate the asymtote. Create a graph, keep in mind it will be a natural asymtote. Natural asymtotes are the same as the number being mathematically added on the end.
• To graph y=3-2^(4-x) find the location of the asymtotes: x=0 and x=1. Plot 4-x to those zero and one variables. Use these variables and create a T chart table. Finish by plotting data and drawing a graph.

Graph Behaviors Overview

• The side of the graph is based on this graph quadrants. If x and y are both positive then it will be on quadrant 1. If y is positivite and x is negative it will go to quadrant 2. If y is negative and x is positive it will go to quadrant 4. Finally is they are both negative it will go towards quadrant 3. Reflacting over with x axis and y , is the say and reflecting it over the origin.
• For logarithmic equations, its recommended to find 0 1 and the value of the chart. Start at the vertical asymtote and follow plotted points; this would be how to graph a basic line following chart points.
• Graph f(x)= log(x-3) +1: The steps require plotting the y values and asymtopes. Determine the y intercept it needs to be 0 1 and value of base chart. Remember when its to 0, its for the graph vertical asymtope. Determine and graph to where the Y axis is located.
• The Domain of logaritithimic, is determined by the lowest point. As determined by this " Domain (3, infinity)

Graph and Inverses Overview

• Y=0 for equations that will exist
• To solve for invers use horizontal asymtote. Use it , rather then a chart. Keep it mind those points is switch in the graph. Graph and vertical line
• Function of typicaly graph will be a vertial asymtope of x=2, function follows exponential formula. Switch lines to e^(x=+2) and see that 2 graph are inverted.