13 times 29 is 337.
Two (2.2308) times.
To express (\log_{13}(145)) using (\log_{13}(2)), (\log_{13}(3)), and (\log_{13}(5)), first factor 145 into its prime components: (145 = 5 \times 29). Then, use the logarithmic property that states (\log_b(mn) = \log_b(m) + \log_b(n)). Since 29 can be expressed as (2 \times 3 \times 5), we have (\log_{13}(145) = \log_{13}(5) + \log_{13}(29)) and can derive (\log_{13}(29)) from the logs of 2 and 3 using their product.
13% of 29= 13% * 29= 0.13 * 29= 3.77
x=29
29 goes into 100 three whole times. After that, there's only room for another 13, so there's no way to stuff the whole 29 in there again.
364
Two (2.2308) times.
13 goes into 29 two times, with a remainder of 3. This can be calculated by dividing 29 by 13, which equals 2 with a remainder of 3. The remainder is the amount left over after dividing the number as many times as possible by the divisor.
To express (\log_{13}(145)) using (\log_{13}(2)), (\log_{13}(3)), and (\log_{13}(5)), first factor 145 into its prime components: (145 = 5 \times 29). Then, use the logarithmic property that states (\log_b(mn) = \log_b(m) + \log_b(n)). Since 29 can be expressed as (2 \times 3 \times 5), we have (\log_{13}(145) = \log_{13}(5) + \log_{13}(29)) and can derive (\log_{13}(29)) from the logs of 2 and 3 using their product.
13% of 29= 13% * 29= 0.13 * 29= 3.77
x=29
29 goes into 100 three whole times. After that, there's only room for another 13, so there's no way to stuff the whole 29 in there again.
Well, let's look at this like a happy little math problem. If we divide 100 by 29, we get about 3 with a remainder. So, 29 goes into 100 about 3 times, with a little bit left over. Remember, there are no mistakes in math, just happy little accidents!
16 + 13 = 29
0.4483
-29
13 13+16=29