Q: What is 3 log x in condensed form?

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[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

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

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

That is the same as log xy.

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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

Using the natural (base e) logs, written as "ln", 3 is eln(3) and 5 is eln(5). Or in base 10, 3=10log(3) and 5=10log(5). Check it out by taking log of both sides: log(3) = log(10log(3)) = log(3) x log(10) =log(3) x 1=log(3).

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If the log of x equals -3 then x = 10-3 or 0.001or 1/1000.

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.

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)

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)

An exponential function can be is of the form f(x) = a*(b^x). Some examples are f1(x) = 3*(10^x), or f2(x) = e^(-2*x). Note that the latter still fits the format, with b = e^(-2). The inverse is the logarithmic function. So for y = f1(x) = 3*(10^x), reverse the x & y, and solve for y:x = 3*(10^y)log(x) = log(3*(10^y)) = log(3) + log(10^y) = log(3) + y*log(10) = y*1 + log(3)y = log(x) - log(3) = log(x/3)The second function: y = e^(-2*x), the inverse is: x = e^(-2*y).ln(x) = ln(e^(-2*y)) = -2*y*ln(e) = -2*y*1y = -ln(x)/2 = ln(x^(-1/2))See related link for an example graph.

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)