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Log x is defined only for x > 0.

The first derivative of log x is 1/x, which, for x > 0 is also > 0

The second derivative of log x = -1/x2 is always negative over the valid domain for x.

Together, these derivatives show that log x is a strictly monotonic increasing function of x and that its rate of increase is always decreasing. Consequently log x is convex.

Q: How do you show log x is convex?

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Log (x^3) = 3 log(x) Log of x to the third power is three times log of x.

Here are a few, note x>0 and y>0 and a&b not = 1 * log (xy) = log(x) + log(y) * log(x/y) = log(x) - log(y) * loga(x) = logb(x)*loga(b) * logb(bn) = n * log(xa) = a*log(x) * logb(b) = 1 * logb(1) = 0

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 base 2 of [x/(x - 23)]

1

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Consider 2 convex functions g(x) and f(x). Using property of second derivative test we have: g(x)'' > 0 and f(x)'' > 0. We are interested in showing that y(x) = g(x) + f(x) is also convex. y(x)'' = c*g(x)'' + k*f(x)'', where c and k are positive numbers (where is no convex function those coeficients after diferentiation will give you negative numbers). Because c, g(x)'', k, f(x)'' are > 0 , we get y(x)'' also >0, hence y(x) is a convex function.

log(x6) = log(x) + log(6) = 0.7782*log(x) log(x6) = 6*log(x)

Log (x^3) = 3 log(x) Log of x to the third power is three times log of x.

log(x) - log(6) = log(15)Add log(6) to each side:log(x) = log(15) + log(6) = log(15 times 6)x = 15 times 6x = 90

Here are a few, note x>0 and y>0 and a&b not = 1 * log (xy) = log(x) + log(y) * log(x/y) = log(x) - log(y) * loga(x) = logb(x)*loga(b) * logb(bn) = n * log(xa) = a*log(x) * logb(b) = 1 * logb(1) = 0

log x + 2 = log 9 log x - log 9 = -2 log (x/9) = -2 x/9 = 10^(-2) x/9 = 1/10^2 x/9 = 1/100 x= 9/100 x=.09

Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.

log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1

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

2*log(15) = log(x) 152 = x; its equivalent logarithmic form is 2 = log15 x (exponents are logarithms) then, it is equivalent to 2log 15 = log x, equivalent to log 152 = log x (the power rule), ... 2 = log15 x 2 = log x/log 15 (using the change-base property) 2log 15 = log x Thus, we can say that 152 = x is equivalent to 2*log(15) = log(x) (equivalents to equivalents are equivalent)

log base 2 of [x/(x - 23)]

5x 12x = 17xx log(5) + x log(12) = x log(17)x [ log(5) + log(12) ] = x log(17)x log(60) = x log(17)x = 0This actually checks. Since anything to the zero power is ' 1 ',50 120 = 1 times 1, or 1and 170 = 1