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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
What multiplication equals 109?
1 x 109
What is 60 percent off of 109 dollars?
109 - 60% = 109 x (1 - (60/100)) = 109 x 0.4 = $43.60
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
What equals 109?
1 x 109, 109 x 1.
What can make 109 by multiplication?
1 x 109, 109 x 1.
What are the prime factors of 109 as a product?
109 and 1 is the only factor pair for 109.
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
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 multiplication equals 109?
1 x 109
Rules of log?
Here are a few, note x>0 and y>0 and a&b not = 1 * log (xy) = log(x) + log(y) * log(x/y) = log(x) - log(y) * loga(x) = logb(x)*loga(b) * logb(bn) = n * log(xa) = a*log(x) * logb(b) = 1 * logb(1) = 0
What times table equals 218?
1 x 218, 2 x 109, 109 x 2, 218 x 1
What is 60 percent off of 109 dollars?
109 - 60% = 109 x (1 - (60/100)) = 109 x 0.4 = $43.60
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)
What time table equals 109?
109 x 1
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
