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Let y = log3 x
⇒ x = 3y
Taking logs to any base you like of both sides gives:
log x = y log 3
⇒ y = log x/log 3
So to calculate log base 3 on a calculator, use either the [log] (common or log to base 10) or [ln] (natural or to base e) function key for the log above, that is use one of:
- [(] [log] [
] [÷] [log] [3] [)] - [(] [ln] [
] [÷] [ln] [3] [)] - [(] [
] [log] [÷] [3] [log] [)] - [(] [
] [ln] [÷] [3] [ln] [)]
Things in square brackets [] represent keys on the calculator; the
Use one of 1 & 1 if your calculator is a more modern one that uses natural representation that looks like maths whereby the calculation is done once you've finished entering it all and the numbers for functions follow them.
Use one of 3 & 4 if your calculator is an older style one that when you press a function key it acts immediately on the number displayed on the screen.
The parentheses (round brackets) are included above so that the whole expression evaluates to log3
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Log base 3 of 81 plus log base 3 of 81?
Log base 3 of 81 is equal to 4, because 3 ^ 4 = 81. Therefore, two times log base 3 of 81 is equal to 2 x 4 = 8.
Which logarithm is equivalent to log base 3 16 - log base 3 2?
log316 - log32 = log38
How do you solve 4 to the power of x equals 128?
If we take the logarithm of both sides, then it is log(4^x) = log(128). Then from logarithm rules, this can be changed to: x*log(4) = log(128), then x = log(128)/log(4). You can punch this into a calculator and get the answer, but what if we use log base 2, we don't need a calculator. So log2(4) = 2, because 2² = 4. And log2(128) = 7, because 2^7 = 128. So we have x = 7/2 = 3.5, then you can check your answer: 4^3.5 = (4^3)*(4^.5). So 4 cubed = 64, and 4 raised to the 1/2 power is the square root of 4, which is 2. So 64 times 2 = 128.
How do I solve the equation e to the power of x equals 2 using logarithms?
When the logarithm is taken of any number to a power the result is that power times the log of the number; so taking logs of both sides gives: e^x = 2 → log(e^x) = log 2 → x log e = log 2 Dividing both sides by log e gives: x = (log 2)/(log e) The value of the logarithm of the base when taken to that base is 1. The logarithms can be taken to any base you like, however, if the base is e (natural logs, written as ln), then ln e = 1 which gives x = (ln 2)/1 = ln 2 This is in fact the definition of a logarithm: the logarithm to a specific base of a number is the power of the base which equals that number. In this case ln 2 is the number x such that e^x = 2. ---------------------------------------------------- This also means that you can calculate logs to any base if you can find logs to a specific base: log (b^x) = y → x log b = log y → x = (log y)/(log b) In other words, the log of a number to a given base, is the log of that number using any [second] base you like divided by the log of the base to the same [second] base. eg log₂ 8 = ln 8 / ln 2 = 2.7094... / 0.6931... = 3 since log₂ 8 = 3 it means 2³ = 8 (which is true).
What does log in algebra mean?
The log or logarithm is the power to which ten needs to be raised to equal a number. Log 10=1 because 10^1=10 Log 100=2 because 10^2=100 Sometimes we use different bases. Like base 2. Then it is what 2 is raised by to get the number. Log "base 2" 8=3 because 2^3=8
