Solving Logarithmic Equations Tuesday, February 9, 2016

Warm-Up

Essential Question How can I use exponential form to help me solve logarithmic equations?

We now know that a logarithm is perhaps best understood as being closely related to an exponential equation. In fact, whenever we get stuck in the problems that follow we will return to this one simple insight. We might even state it as a simple rule. We now know that a logarithm is perhaps best understood as being closely related to an exponential equation. In fact, whenever we get stuck in the problems that follow we will return to this one simple insight. We might even state it as a simple rule. Solving equations with logarithms:

When working with logarithms, if ever you get “stuck”, try rewriting the problem in exponential form. Conversely, when working with exponential expressions, if ever you get “stuck”, try rewriting the problem in logarithmic form.

Solution: Let’s rewrite the problem in exponential form. We’re finished, x = 36

Solution: Rewrite the problem in exponential form. Alternate: If you can evaluate the log, you will already have the value of the y! log 5 (1/5 2 ) = log = -2

Example 3 Try setting this up like this: Solution: Rewrite in exponential form. Solve for x… log x 54 = 4 x 4 = 54 Now use your reciprocal power to solve. (x 4 ) 1/4 = (54) 1/4 x = (54) 1/4 So x ≈

Example 4 Solution: Rewrite in exponential form. Solve for x… log 3 (x+4) = = x + 4 Simplify the power expression. 81 = x + 4 x = 77 Subtract 4 to finish…

Example 5: More complicated expressions KEY KNOWLEDGE: You ALWAYS have to isolate the log expression before you change forms. Rewrite in exponential form. Remember common log is base 10. Solve for x… 4 log (x+1) = = x + 1 Evaluate the power. 100 = x + 1 x = 99 Subtract 1 to finish… Divide by the 4… log (x+1) = 2

Example 6: More complicated expressions KEY KNOWLEDGE: You ALWAYS have to isolate the log expression before you change forms. Rewrite in exponential form.. Solve for x… 0.5 log 2 (3x+1) – 6 = –4 2 4 = 3x + 1 Evaluate the power. 16 = 3x + 1 x = 5 Subtract 1, then divide by 3 to finish… Add 6… 0.5 log 2 (3x+1) = 2 log 2 (3x+1) = 4 Multiply by 2 (or divide by 0.5) 15 = 3x

Finally, we want to take a look at the Property of Equality for Logarithmic Functions. Basically, with logarithmic functions, if the bases match on both sides of the equal sign, then simply set the arguments equal.

Example 7 Solution: Since the bases are both ‘3’ we simply set the arguments equal. Is the –3 a valid solution or is it extraneous?? Note: If I plug -3 into the original, it causes me to take the log of a negative number!!! Therefore, it is NOT valid. Our answer is NO SOLUTION…

Example 8 Solution: Since the bases are both ‘8’ we simply set the arguments equal. Factor the quadratic and solve continued on the next page

Example 8 (cont) Solution: It appears that we have 2 solutions here. If we take a closer look at the definition of a logarithm however, we will see that not only must we use positive bases, but also we see that the arguments must be positive as well. Therefore, -2 is not a solution. Each side becomes the log 8 (-10) !!!!! So that answer is extraneous and has to be thrown out. Only x = 7 works in the original equation.

Example 9: Equations with more than one log (Using our Condensing Skills) KEY KNOWLEDGE: You ALWAYS have to isolate the log expression before you change forms. You may have to CONDENSE it! Now solve... Subtract 3x. Solve for x… log 2 (3x + 8) = log log 2 x 8 = x x = 8 Condense the right side … log 2 (3x+8) = log 2 (4x) (Product Property) 3x + 8 = 4x Property of Equality…

Example 10: Equations with more than one log (Using our Condensing Skills) KEY KNOWLEDGE: You ALWAYS have to isolate the log expression before you change forms. You may have to CONDENSE it! Now solve... Divide by 24 and simplify. Solve for x… log 2 (3x) + log 2 (8) = 4 x = 16/24 x = 2/3 Condense the left side … log 2 (24x) = 4 (Product Property… 3x*8 = 24x) 2 4 = 24x Change into exponential form… Evaluate the power… 16 = 24x

HOMEWORK: