To find the antilog of a negative number using a log table, first, add the absolute value of the negative number to the characteristic of the log table. Next, locate the resulting number in the log table to find the corresponding mantissa. Finally, take the antilog of the mantissa to get the final answer. Remember to consider the negative sign of the original number when determining the final result.
log on to
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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
You calculate a log, you do not solve a log!
no
log(36) = 1.5563To solve this problem without using a scientific calculator, factor 36 into 2*2*3*3, and use the formula:log(a*b) = log(a) + log(b)So, in this case:log(36) = log(2) + log(2) + log(3) + log(3) = 0.3010 + 0.3010 + 0.4772 + 0.4772 = 1.5564 (slight rounding error)
log(10) 12 = 1.07918 Then the antilog is 12 = 10^(1.07918) You must specify the base to which to logarithm is functioning. Different log bases will give different answers.
To find anti log of a number enter the number as the exponent of 10.
The anti-log of 12.34 is the inverse operation of taking the logarithm of a number. In this case, the anti-log of 12.34 is equal to 10^12.34, which is approximately 2511886431. A logarithm is the power to which a base must be raised to produce a given number, so the anti-log reverses this operation to find the original number.
You Log Off
log(24.6992) = 1.392682887 Now the anti-log. 101.392682887 = 24.6992 ---------------
Without antilog tables or a scientific calculator you cannot. Antilog(x) is usually 10x or ex and that is not simple to calculate.
Navy Log - 1955 Night Landing 1-36 was released on: USA: 29 May 1956
You have, y = 6 + log x anti log of it, 10y = (106) x
Adam-12 - 1968 Log 36 Man Between 3-18 was released on: USA: 4 February 1971
You can do it directly on a calculator and get 1.2676506002282e+30. Or you can use logs like y=n^5 and log y=5*log(1048576)=30.102999566398. Look up the anti-log for the answer.