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How do you sole X when logx-3 plus logx-2 equals log2x plus 24?
We can't see the parentheses, and there are at least two ways to read this.Here are solutions for the most likely two:--------------------------------------------------------log(x) - 3 + log(x) - 2 = log(2x) + 24Add 5 to each side:log(x) + log(x) = log(2x) + 29Subtract log(2x) from each side:log(x) + log(x) - log(2x) = 29Combine the logs on the left side, and massage:log( x2/2x ) = log( x/2 ) = 29Take the antilog of each side:x/2 = 1029Multiply each side by 2:x = 2 x 1029------------------------------------------------log(x - 3) + log(x - 2) = log(2x + 24)Combine logs on the left side:log[ (x-3) (x-2) ] = log(2x + 24)Take antilog of each side:(x-3) (x-2) = 2x + 24Expand the left side:x2 - 5x +6 = 2x + 24Subtract (2x+24) from each side:x2 - 7x - 18 = 0Factor:(x - 9) (x + 2) = 0Whence:x = 9x = -2We have to discard the solution [ x = -2 ] because one term in the equationis log(x-2).If 'x' were -2 then we'd have log(-4) but negative numbers don't have logs.
Solve for x. 2log2x plus log21 equals log24?
I will assume that the "2" after each log is a subscript (indicating log to the base 2). Basically, you must use the well-known logarithmic identities, a log b = log ba, and log a + log b = log ab. 2 log2x + log21 = log24 log2x2 + log21 = log24 log2x2(1) = log24 log2x2 = log24 Take antilogs on both sides: x2 = 4 In this last equation, x is either 2 or -2. However, negative numbers are not appropriate for the original equation (assuming real numbers), so the only solution is 2. For safety, this should be checked in the original equation; I'll leave that part to you.
How to differentiate ln x 2?
Let y = ln(x2), where ln is log to base e. Using our log laws, y = 2 ln (x) Therefore dy/dx = 2 . 1/x = 2/x Another way to do this is to use the chain rule (for differentiation) Let y = ln(x2), where ln is log to base e. Let u = x2, therefore du/dx = 2x y = ln u, therefore dy/du = 1/u Putting it altogether gives: dy/dx = dy/du . du/dx = 1/u . 2x = 2x/x2 = 2/x
What is x2 plus x2 plus x2?
x2 + x2 + x2 = (1 + 1 + 1)x2 = 3x2
What is 3x4 plus 5x3 plus x2 - 5 divided by x 2?
3x4 plus 5x3 plus x2 - 5 divided by x 2 =[(3x4) + (5x3) + (x2 - 5)]/x2 =(12 + 15 + x2 -5)/x2 =(27 - 5 + x2)/x2 =(22 + x2)/x2
Log9x plus log x equals 4log10?
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
What does x equals in 2 log x equals log 8?
2 log(x) = log(8)log(x2) = log(8)x2 = 8x = sqrt(8) = 2.82843 (rounded)Note that only the positive square root of 8 can serve as a solution to thegiven equation, since there's no such thing as the log of a negative number.
Why is the answer for log x squared equals 2 different than 2log x equals 2?
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)
Whats the answer to x2 plus 3x plus 1?
The roots are -2.6180 and -0.3820
How do you sole X when logx-3 plus logx-2 equals log2x plus 24?
We can't see the parentheses, and there are at least two ways to read this.Here are solutions for the most likely two:--------------------------------------------------------log(x) - 3 + log(x) - 2 = log(2x) + 24Add 5 to each side:log(x) + log(x) = log(2x) + 29Subtract log(2x) from each side:log(x) + log(x) - log(2x) = 29Combine the logs on the left side, and massage:log( x2/2x ) = log( x/2 ) = 29Take the antilog of each side:x/2 = 1029Multiply each side by 2:x = 2 x 1029------------------------------------------------log(x - 3) + log(x - 2) = log(2x + 24)Combine logs on the left side:log[ (x-3) (x-2) ] = log(2x + 24)Take antilog of each side:(x-3) (x-2) = 2x + 24Expand the left side:x2 - 5x +6 = 2x + 24Subtract (2x+24) from each side:x2 - 7x - 18 = 0Factor:(x - 9) (x + 2) = 0Whence:x = 9x = -2We have to discard the solution [ x = -2 ] because one term in the equationis log(x-2).If 'x' were -2 then we'd have log(-4) but negative numbers don't have logs.
How do you show log x is convex?
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.
Solve for x. 2log2x plus log21 equals log24?
I will assume that the "2" after each log is a subscript (indicating log to the base 2). Basically, you must use the well-known logarithmic identities, a log b = log ba, and log a + log b = log ab. 2 log2x + log21 = log24 log2x2 + log21 = log24 log2x2(1) = log24 log2x2 = log24 Take antilogs on both sides: x2 = 4 In this last equation, x is either 2 or -2. However, negative numbers are not appropriate for the original equation (assuming real numbers), so the only solution is 2. For safety, this should be checked in the original equation; I'll leave that part to you.
How do you solve sin x equals 6?
First, take the inverse sine of both sides of the equation. That gives you x = sin-1(6), which is sadly undefined...in reality, but who needs that! It can be proven that sin-1(x) = -i*log[i*x + √(1-x2)] So in this case: = -i*log[i*6 + √(1-36)] = -i*log[6*i + √(-35)] = -i*log(11.916*i) = 1.57 - 2.48*i
How to differentiate ln x 2?
Let y = ln(x2), where ln is log to base e. Using our log laws, y = 2 ln (x) Therefore dy/dx = 2 . 1/x = 2/x Another way to do this is to use the chain rule (for differentiation) Let y = ln(x2), where ln is log to base e. Let u = x2, therefore du/dx = 2x y = ln u, therefore dy/du = 1/u Putting it altogether gives: dy/dx = dy/du . du/dx = 1/u . 2x = 2x/x2 = 2/x
What is x2 plus x2 plus x2?
x2 + x2 + x2 = (1 + 1 + 1)x2 = 3x2
Whats the answer to Factor x3-3x2 plus x-3?
x3 - 3x2 + x - 3 = (x - 3)(x2 + 1)
What is X2 plus X2?
x2 + x2 = 2x2
