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How do you isolate x in this equation 1 equals leftbracket 1 plus x right bracket exponent 5?
with something called logarithms. So 1 = (1 + x)^5 log 1 = log ((1+x)^5) log 1 = 5 x log (1 +x) but log 1 = 0 therefore 0 = 5 x log(1+x) divide both sides by 5 and you get 0 = log (1+x) we know that log 1 = 0, therefore 1+ x = 1 and so x = 0
5 to the x power times 12 to the x power equals 17 to the x power what is x?
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
What is the answer to Log x plus 4 minus log6 equals 1?
log(x) + 4 - log(6) = 1 so log(x) + 4 + log(1/6) = 1 Take exponents to the base 10 and remember that 10log(x) = x: x * 104 * 1/6 = 10 x = 6/1000 or 0.006
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
X minus 1 plus 2 plus log x equals 3. find x?
Original Statement:x - 1 + 2 + log(x) = 3Simplify:x + 1 + log(x) = 3Subtract 1:x + log(x) = 2Lambert W-Function:x = (W(100*ln(10))/(ln(10)) = 1.7555794993... (rounded up).This considered log(x) to be base 10 log (x).
