0
No, but logx(xx) = x
However for all bases:
- logx 1 = 0
- logx x = 1
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When does log p X equals 0?
When X is 1, regardless of the base p.
How do you solve log x plus log 8 equals 1?
log(x)+log(8)=1 log(8x)=1 8x=e x=e/8 You're welcome. e is the irrational number 2.7....... Often log refers to base 10 and ln refers to base e, so the answer could be x=10/8
Solve ln y equals xln e?
ln is the natural logarithm. That is it is defined as log base e. As we all know from school, log base 10 of 10 = 1 just as log base 3 of 3 = 1, so, likewise, log base e of e = 1 and 1.x = x. so we have ln y = x. Relace ln with log base e, and you should get y = ex
What is the answer to Log x plus 4 minus log6 equals 1?
log(x) + 4 - log(6) = 1 so log(x) + 4 + log(1/6) = 1 Take exponents to the base 10 and remember that 10log(x) = x: x * 104 * 1/6 = 10 x = 6/1000 or 0.006
X minus 1 plus 2 plus log x equals 3. find x?
Original Statement:x - 1 + 2 + log(x) = 3Simplify:x + 1 + log(x) = 3Subtract 1:x + log(x) = 2Lambert W-Function:x = (W(100*ln(10))/(ln(10)) = 1.7555794993... (rounded up).This considered log(x) to be base 10 log (x).
When does log p X equals 0?
When X is 1, regardless of the base p.
How do you solve log x plus log 8 equals 1?
log(x)+log(8)=1 log(8x)=1 8x=e x=e/8 You're welcome. e is the irrational number 2.7....... Often log refers to base 10 and ln refers to base e, so the answer could be x=10/8
What is the answer to Log x plus 4 minus log6 equals 1?
log(x) + 4 - log(6) = 1 so log(x) + 4 + log(1/6) = 1 Take exponents to the base 10 and remember that 10log(x) = x: x * 104 * 1/6 = 10 x = 6/1000 or 0.006
Solve ln y equals xln e?
ln is the natural logarithm. That is it is defined as log base e. As we all know from school, log base 10 of 10 = 1 just as log base 3 of 3 = 1, so, likewise, log base e of e = 1 and 1.x = x. so we have ln y = x. Relace ln with log base e, and you should get y = ex
The value of log log4log4x equals 1 then x equals?
the value of log (log4(log4x)))=1 then x=
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)
If y equals 10 then what is then what is y equals log x?
y = 10 y = log x (the base of the log is 10, common logarithm) 10 = log x so that, 10^10 = x 10,000,000,000 = x
X minus 1 plus 2 plus log x equals 3. find x?
Original Statement:x - 1 + 2 + log(x) = 3Simplify:x + 1 + log(x) = 3Subtract 1:x + log(x) = 2Lambert W-Function:x = (W(100*ln(10))/(ln(10)) = 1.7555794993... (rounded up).This considered log(x) to be base 10 log (x).
What is x if log x equals -3?
If the log of x equals -3 then x = 10-3 or 0.001or 1/1000.
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).
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
Differentiate log x?
The derivative of ln x, the natural logarithm, is 1/x.Otherwise, given the identity logbx = log(x)/log(b), we know that the derivative of logbx = 1/(x*log b).ProofThe derivative of ln x follows quickly once we know that the derivative of ex is itself. Let y = ln x (we're interested in knowing dy/dx)Then ey = xDifferentiate both sides to get ey dy/dx = 1Substitute ey = x to get x dy/dx = 1, or dy/dx = 1/x.Differentiation of log (base 10) xlog (base 10) x= log (base e) x * log (base 10) ed/dx [ log (base 10) x ]= d/dx [ log (base e) x * log (base 10) e ]= [log(base 10) e] / x= 1 / x ln(10)
