Use the identity log(ab) = log a + log b to combine the logarithms on the left side into a single term. Then take antilogarithms (just take the log away) on both sides.
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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
True
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Use the identity log(ab) = log a + log b to combine the logarithms on the left side into a single term. Then take antilogarithms (just take the log away) on both sides.
No. Insert the word "minus" in place of the word "plus", and you'll have a true statement.
The explanation and answer to the following math equation to find x -0.3 plus 5-5 log (d) equals a plus 5-5 log 4 (d) is -5 log(d)+x+4.7 = a-5 log(4 d)+5. The solution is x = a-6.63147.
log(-3,2.3914850069628266…i) = 1.26ie. (-3)1÷1.26=2.3914850069628266…i
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "equals".
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
True
Unfortunately, limitations of the browser used by WA means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals" etc.
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
G(x) = log(2x) + 2, obviously!
2