log3x - 2logx3 = 1
so
1/logx3 - 2logx3 = 1
Let y = logx3
then 1/y - 2y = 1
multiply through by y: 1 - 2y2 = y
that is, 2y2 + y - 1 = 0
which factorises to (y + 1)(2y -1) = 0
so that y = -1 or y = 1/2
y = -1 implies logx3 = -1 so that x-1 = 3 ie 1/x = 3 or x= 1/3
y = 1/2 implies logx3 = 1/2 so that x1/2 = 3 or x = 32 = 9
log x2 = 2 is the same as 2 log x = 2 (from the properties of logarithms), and this is true for x = 10, because log x2 = 2 2 log x = 2 log x = 1 log10 x = 1 x = 101 x = 10 (check)
log(9x) + log(x) = 4log(10)log(9) + log(x) + log(x) = 4log(10)2log(x) = 4log(10) - log(9)log(x2) = log(104) - log(9)log(x2) = log(104/9)x2 = 104/9x = 102/3x = 33 and 1/3
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).
2x = 0.5: This is like asking for the logarithm of 0.5, to the base 2. A scientific calculator normally has logarithms for base 10 and base e, but not for other bases. However, you can calculate this is log(0.5) / log(2). It doesn't matter what base you use for your logarithms, just keep it consistent. For example, with base 10, log(0.5) / log(2) = -0.301 / 0.301 = -1.
If the log of x equals -3 then x = 10-3 or 0.001or 1/1000.
d/dx (2 log(1) + x) = 1
log x2 = 2 is the same as 2 log x = 2 (from the properties of logarithms), and this is true for x = 10, because log x2 = 2 2 log x = 2 log x = 1 log10 x = 1 x = 101 x = 10 (check)
log(9x) + log(x) = 4log(10)log(9) + log(x) + log(x) = 4log(10)2log(x) = 4log(10) - log(9)log(x2) = log(104) - log(9)log(x2) = log(104/9)x2 = 104/9x = 102/3x = 33 and 1/3
When X is 1, regardless of the base p.
The logarithm of 0.5 to the base 2, written as log₂(0.5), is equal to -1. This is because 2 raised to the power of -1 equals 0.5 (i.e., 2^(-1) = 1/2 = 0.5). Thus, log₂(0.5) = -1.
Your expression is this... 2log(1/2)3 - log(1/2)(2x-3) = 1 Now just plug the numbers into the calculator to get... -1.806 + 0.602x - 0.903 = 1 Combine like terms... 0.602x = 3.709 Find x... x=6.16
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
No, but logx(xx) = xHowever for all bases:logx 1 = 0logx x = 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
The expression (3 \log 10) can be simplified using the properties of logarithms. Since (\log 10) in base 10 equals 1, we have (3 \log 10 = 3 \times 1 = 3). Therefore, (3 \log 10 = 3).
the value of log (log4(log4x)))=1 then x=
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)