*First off if we assume this log to be base 10
next we can use the product rule (d/dx (3)*logx+d/dx(logx)*3)
1.derivative of a constant is zero
so that gives us 0*logx as our first term (simplifies to zero)
next we have to differentiate logx
that gives us 3*(1/xln(10))
so that leaves 0logx+3*(1/xln(10))
simplify...... 3/xln(10)
Log (x^3) = 3 log(x) Log of x to the third power is three times log of x.
1
If the log of x equals -3 then x = 10-3 or 0.001or 1/1000.
x = 3*log8 = log(83) = log(512) = 2.7093 (approx)
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
Log (x^3) = 3 log(x) Log of x to the third power is three times log of x.
[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
1
If the log of x equals -3 then x = 10-3 or 0.001or 1/1000.
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)
x = 3*log8 = log(83) = log(512) = 2.7093 (approx)
3x = 18Take the logarithm of each side:x log(3) = log(18)Divide each side by log(3):x = log(18) / log(3) = 1.25527 / 0.47712x = 2.63093 (rounded)
log(1) = 0log(x*y) = log(x) + log(y)If logb(x) = y then x = by.
log(x) = 3x = 10log(x) = 103 = 1,000
log (3 x 66) = log 3 + log 66
(1+x)10 = 310 log(1+x) = log(3)log(1+x) = 0.1 log(3)(1+x) = 10[0.1 log(3)] = 1.116123x = .116123 = 11.61 percent
3 sec23x