It's time for our master-class before facing off with Expo and his minions. Log sends us off to an old, musty library with stack after stack of books. They've got titles like Ye Olde Mathematical Beasts and Logarithmica Adeptus. Whenever you open up one of the ancient books, dust puffs out all over your face. Don't sneeze, though, that'll just stir it up even more.
Looking through them we find some properties of logarithms and exponentials that will help you out. Some we've already met, some are fresh and new.
Sum of Logs: log_{b}xy = log_{b}x + log_{b}y
For all of these properties, looking at them in their exponential form is a big help. Let's do that:
e = log_{b}xy is the same as b^{e} = xy
f + g = log_{b} x + log_{b} y is the same as b^{f} × b^{g} = xy
Remember that multiplying two numbers with the same base together will cause them to merge together and add the exponents together. For example 3 × 3 = 3^{1} + 3^{1} = 3^{2}, so b^{f} + b^{g} = b^{e} or b^{f + g}.
Difference of Logs:
This one is similar to the previous property, but just in reverse gear. If two numbers with the same base are divided, their exponents are subtracted from each other.
Exponent in Log: log_{b} x^{n} = n log_{b} x
Whoa, how'd that exponent just decide to go outside the log? Expo just doesn't care 'bout anyone now, it seems. Zip, zoom, fly—it's time for exponent mode.
b^{e} = x^{n}
Imagine the following situation: 10^{2} = 100, (n = 1).
If we cube each side, this is what would go down:
(10^{2})^{3} = 100^{3}
The power of 3 on the right side is n in both the logarithmic and exponential form. All you have to remember here is that when you raise an exponent to another exponent, you can multiply them:
10^{2 × 3} = 100^{3}
10^{6} = 100^{3}
Demonstrate the sum of logs by expanding log 100 and solving, then demonstrate the log in exponent property by expanding.
There are two other ways to represent log 100: log 10 × 10 or log 10^{2}.
If we expand the first way, we get log 10 + log 10 = 1 + 1 = 2. log 100 is also 2.
If we expand the second way, we get 2 log 10 = 2 × 1 = 2. No sweat.
3 log_{4} 16x^{5}
First let's pull that exponent down from its pedestal:
3(5) log_{4} 16x = 15 log_{4} 16x
Then pull the logs apart, sure to hold onto the coefficient:
15 log_{4} 16 + 15 log_{4} x
Simplify the first log, because
4^{2} = 16: 15(2) + 15 log_{4} x = 30 + 15 log_{4} x
There's your answer:
30 + 15 log_{4} x.
Log of 1: log_{b} 1 = 0
Essentially all this says is, "what exponent do we raise a base to so we get an answer of 1?" Because the base cannot be 1, that only leaves one answer: 0. Absolutely any base raised to the power of 0 is 1. The base could be 1,000,000, but raising it to the 0 power would still give us 1. How mystical.
Equal Base: log_{b} b = 1
Again, let's put on our exponential-form goggles. (Now you look like you're on a secret mission. So classy.)
b^{e} = b
The logarithm answers the question, "What's the exponent?" If we need the two sides to equal, the only exponent that would give a correct answer would be one. It'd work out just to be b = b.
Rewrite the following logarithmic functions into a single log, simplifying where possible, using the properties you've learned.
y = log_{5} 20x^{2} + 1/2 log_{5} 25x – log_{5} 5
Whoa boy. That's a Moby Dick-sized whale of a problem.
Let's go from left to right. First, pull out the exponent in the left log:
2 log_{5} x
Split the middle log into a sum:
Now that you've got all the skills to beat Expo at his own game, taking him down should be a cinch.
Expand the following logarithmic equation: log (16x^{2}y^{2})^{1/3} |
Combine the following logarithmic equations into one: |
Simplify the following exponential function to logarithmic form, then simplify irrationals to three decimal places and convert back to exponential form: y = 10^{2x}7^{x} |
Is y = log_{2} 4x equivalent to 2^{y – 2} = x If so, show how. |
Expand the following logarithmic equation: |
Expand the following equation: log 27x^{2} |
Expand the following log equation, and write it using only base-10 logarithmic functions. Solve irrationals to 3 decimal places: y = ln (4^{e})(3x + 1) |
Is If so, show how. |
Can the following expression be simplified: y = (log x)(log 2)? |
Combine the following log functions into one function: log 4 + log (x – 3) + log (2x + 4)
Is the following expansion of a logarithmic function correct: log (9x + 4) = log 9x + log 4
Is y = log_{4} 16 equivalent to y = 2 log_{4} 8?
Combine the following log functions into one function: log 7 – log 8x – log (4x + 6)
Combine the following log functions into one function: log (x^{2}) + log y – log y^{2}
Simplify the following equation: xe^{4x}e^{-7x} = y
Expand the following log equation, and write it using only base-10 logarithmic functions: y = log_{7} (5x – 4)(2x + 3)
Expand the following log function: log_{5} 4xy
Expand the following log function, then simplify irrationals to three decimal places:
Simplify the following log function to a form without exponents, then change to base 10: y = log_{7} 100x^{2}
Convert the following exponential equation to natural logarithmic form, then simplify irrationals to three decimal places: y = e^{x}4^{x}
Simplify the following log function: log (x^{2} + 4x + 4)
Simplify the following log function and solve for z: 4 = log x^{2} + log y + log z
Expand the following log function: ln (4x^{2}y/z^{1/2})