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*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)

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Q: Differentiate 3 log x

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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

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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.

3 sec23x

x = 3*log8 = log(83) = log(512) = 2.7093 (approx)

If log(x) = y then log(x3) = 3*log(x) = 3*y so that x3 = antilog(3*y) So, to find the cibe of x 1) find log x 2) multiply it by 3 3) take the antilog of the result.

log(x) = 3x = 10log(x) = 103 = 1,000

log (3 x 66) = log 3 + log 66

log(1) = 0log(x*y) = log(x) + log(y)If logb(x) = y then x = by.

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)

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)

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