2n=3.91×1012 2^n = 3.91 \times 10^{12}

Given some equation like the one above, with a base and the result after raising it to a power, how do we find the power that was used? Essentially, we are taking the inverse of the exponential function. This inverse function is a logarithm written as "log".

Calculating the logarithm log2(32) \log_2(32) is akin to asking: 2 2 raised to what power is 32? 32? Since 25=32, 2^5 = 32 , log2(32)=5. \log_2(32) = 5 .

What is log6(36)? \log_6(36) ?



Which of these two numbers is larger?

log10(1000) and log2(9) \log_{10}(1000) \quad \text{ and } \quad \log_2(9)

Remember, you can read logb(x) \log_b(x) as "b b raised to what power is x? x ?"



Since exponents don't need to be integers...

43.5=128 4^{3.5} = 128

...the results of logarithms don't need to be either!

log4(128)=3.5 \log_{4}(128) = 3.5

We can think of the logarithms that come out to be integers as "reference points".

log22=1log24=2log28=3log216=4 \log_{2}{2} = 1 \quad \log_{2}{4} = 2 \quad \log_{2}{8} = 3 \quad \log_{2}{16} = 4

log27 \log_2{7} is...



log10(1000)=3 \log_{10}(1000) = 3

log10(49323)4.69 \log_{10}(49323) \approx 4.69

log10(333333)5.52 \log_{10}(333333) \approx 5.52

log10(100000)=5 \log_{10}(100000) = 5

Which statement is true? ((Assume you are working with numbers in base 10.)10.)



log2(12) \log_2(12) is...



We're going to finish our tour of exponential and logarithm functions by going back into space on a journey to the edge of the universe. We'll use something called a logarithmic scale to help us out, and establish the framework for the mathematics that starts our Pre-Calculus course. Read on!



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