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You cannot solve log x- 2 unless (i) log x - 2 is equal to some number or (ii) x is equal to some number.

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Q: How do you solve log x - 2?
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Related questions

How do you solve log x plus 2 equals log 9?

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


How do you solve for X in Log x plus 9 - Log x equals 2?

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Solve for x log 9 x equals 2?

log9(x)=2 x=9^2 x=81


How do you solve log base 2 of x - 3 log base 2 of 5 equals 2 log base 2 of 10?

[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


How do you solve 2logx plus 3logx equals 10?

2 log(x) + 3 log(x) = 105 log(x) = 10log(x) = 10/5 = 210log(x) = (10)2x = 100


How do you solve 3 log of 8 equals x?

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


How do you solve 3 to the power of negative 2x plus 2 equals 81?

3^(-2x + 2) = 81? log(3^(-2x + 2)) = log(81) (-2x+2)log(3) = log(81) -2x = log(81)/log(3) - 2 x = (-1/2)(log(81)/log(3)) + 1


Log x plus log 2 equals log 2?

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


Log base 2 of x - log base 2 of x-23?

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


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)


How do you solve a log?

You calculate a log, you do not solve a log!


What is the logarithmic form of 15 squared equals x?

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)