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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
How do you key in log 2 on your TI 86 calculator?
Well, darling, to calculate log base 2 on your TI-86 calculator, you simply press the "LOG" button, then type in "2" and hit enter. Voila! You've got your answer. Now go forth and conquer those logarithms like the math boss you are.
What is the log of 125?
What 'logarithm base are you using. If Base '10' per calculator The log(10)125 = 2.09691 However, You can use logs to any base So if we use base '5' Then log(5)125 = 3 Because 125 = 5^3
How do you use the calculator to find the cube root?
Depends on your calculator. If you have "raise to the power" then use "raise to the power 1/3". If not, try logs: either logs to base 10 or logs to base e will do: find the log, divide it by 3, then find the antilog. For base e, (log sometimes written "ln" meaning "natural log") the antilog is just the exponential : " ex ".
What is log base 5of 125?
log(5)125 = log(5) 5^(3) = 3log(5) 5 = 3 (1) = 3 Remember for any log base if the coefficient is the same as the base then the answer is '1' Hence log(10)10 = 1 log(a) a = 1 et.seq., You can convert the log base '5' , to log base '10' for ease of the calculator. Log(5)125 = log(10)125/log(10)5 Hence log(5)125 = log(10) 5^(3) / log(10)5 => log(5)125 = 3log(10)5 / log(10)5 Cancel down by 'log(10)5'. Hence log(5)125 = 3 NB one of the factors of 'log' is log(a) a^(n) The index number of 'n' can be moved to be a coefficient of the 'log'. Hence log(a) a^(n) = n*log(a)a Hope that helps!!!!!
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 answer log 8 plus log 3 the base is assumed to be 10?
log base 10 of 24. Use your calculator. log(24)Thanks, but i mean after you get to log 10 of 24 it looks like this24=10^x how do I figure this i meanType in the "log(" button, then 24 if you're using a graphing calculator.Type in 24 then "log" if you're using a small scientific calculator.Spreadsheet programs can do it as well. Type this:=log(24)in a cell and press the Enter key.
What is log base 3 of (x plus 1) log base 2 of (x-1)?
The browser which is used for posting questions is almost totally useless for mathematical questions since it blocks most symbols.I am assuming that your question is about log base 3 of (x plus 1) plus log base 2 of (x-1).{log[(x + 1)^log2} + {log[(x - 1)^log3}/log(3^log2) where all the logs are to the same base - whichever you want. The denominator can also be written as log(3^log2)This can be simplified (?) to log{[(x + 1)^log2*(x - 1)^log3}/log(3^log2).As mentioned above, the expression can be to any base and so the expression becomesin base 2: log{[(x + 1)*(x - 1)^log3}/log(3) andin base 3: log{[(x + 1)^log2*(x - 1)}/log(2)
How do you find antilog of negative number?
To find the antilog of a negative number, you can use the formula antilog(x) = 10^x, where x is the negative number. The antilog of a negative number represents the inverse operation of finding the power of 10 that results in the negative number.
How do you solve log base 2 of x - 3 log base 2 of 5 equals 2 log base 2 of 10?
[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5
How do you write expressions in exponential form of the numbers 3 and 5?
Using the natural (base e) logs, written as "ln", 3 is eln(3) and 5 is eln(5). Or in base 10, 3=10log(3) and 5=10log(5). Check it out by taking log of both sides: log(3) = log(10log(3)) = log(3) x log(10) =log(3) x 1=log(3).
How do we change from logarithmic form to exponential form if it doesn't have a exponent?
ln x = 3 becomes x = e3 for natural logarithms e is the base the side opposite the log side becomes the exponent. ln 3 = x ... use a calculator or log table to find the value of x same for logs of any other base.
