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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).
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
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
How do you solve log x plus log 8 equals 1?
log(x)+log(8)=1 log(8x)=1 8x=e x=e/8 You're welcome. e is the irrational number 2.7....... Often log refers to base 10 and ln refers to base e, so the answer could be x=10/8
Log x plus 7 equals 1 plus Log x-1?
logx +7=1+log(x-1) 6=log(x-1)-logx 6=log[(x-1)/x] 10^6=(x-1)/x 1,000,000x=x-1 999,999x=-1 x=-1/999,999
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).
What is log base 3 of (x plus 1) log base 2 of (x-1)?
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)
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
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
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
What is the derivative of 2log 1 plus x?
d/dx (2 log(1) + x) = 1
How do you solve for X in Log x plus 9 - Log x equals 2?
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How do you solve log x plus log 8 equals 1?
log(x)+log(8)=1 log(8x)=1 8x=e x=e/8 You're welcome. e is the irrational number 2.7....... Often log refers to base 10 and ln refers to base e, so the answer could be x=10/8
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
Can anyone solve for x log 4 X plus 16 plus log 4 X plus 4 equals 3?
log4(x) +16 +log4(x) +4=32log4(x)=-17log4(x)=-17/2x=4^(-17/2)=========================Since the parentheses have been lost from the question,it could easily be interpreted this way instead, (as well asa few others):x log(4x) + 16 + log(4x) + 4 = 3(x + 1) log(4x) = -174x = 10-17/(x+1)4x = the (x+1)th root of 10-17Come on back and solve that one for us.
