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.
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
log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
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
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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.
2 log(x) + 3 log(x) = 105 log(x) = 10log(x) = 10/5 = 210log(x) = (10)2x = 100
3^(-2x + 2) = 81? log(3^(-2x + 2)) = log(81) (-2x+2)log(3) = log(81) -2x = log(81)/log(3) - 2 x = (-1/2)(log(81)/log(3)) + 1
You calculate a log, you do not solve a log!
x = 3*log8 = log(83) = log(512) = 2.7093 (approx)
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
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
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log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
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
If log(Kf) = 5.167 then Kf = 105.167 = 146,983 (approx).
"Log" is not a normal variable, it stands for the logarithm function.log (a.b)=log a+log blog(a/b)=log a-log blog (a)^n= n log a