log3(x)=4 x=3^4 x=81
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
log37 - log3x = 4 log3(7/x) = 4 7/x = 34 = 81 x = 7/81
log3 + logx=4 log(3x)=4 3x=10^4 x=10,000/3
Molarity of hydrogen solution equals 2.3 X 10^-4 -log(2.3 X 10^-4) = 3.6 pH
log3(x)=4 x=3^4 x=81
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
log37 - log3x = 4 log3(7/x) = 4 7/x = 34 = 81 x = 7/81
If log10X = 4, then x = 104 = 10 000 or 10,000 depending on how you prefer to write it.
log3 + logx=4 log(3x)=4 3x=10^4 x=10,000/3
If the log of x equals -3 then x = 10-3 or 0.001or 1/1000.
Molarity of hydrogen solution equals 2.3 X 10^-4 -log(2.3 X 10^-4) = 3.6 pH
the value of log (log4(log4x)))=1 then x=
y = 10 y = log x (the base of the log is 10, common logarithm) 10 = log x so that, 10^10 = x 10,000,000,000 = x
log(x) = 3x = 10log(x) = 103 = 1,000
-log(9.40 X 10^-4) = 3 pH
Without brackets, there are many ambiguities in the question. 4 = 2/3*log(x) - 0.9 4.9*3/2 = logx 7.35 = logx so x = 107.35 However, the original equation could also have referred to 4 = 2/3*log(x-0.9) or 4 = 2/[3*log(x)] - 0.9 or 4 = 2/[3*log(x-0.9)]