250(x) = 400,000 Use logs to base '10' Hence log250^(x) = log400,000 xlog250 = log 400,000 Notirce how the power (x) becomes the coefficient. This is oerfectly correct under 'log' rules. x = log 400,000 / log 250 (NOT log(250/400,000). x = 5.60206 / 2.39794 x = 2.33619689..... ~ 2.33619 to 5 d.p.
Use logarithms to base '10. Hence log(10) a^(1/50) Fiftieth root) and log (10) a^(1/2) second root ( square root) Under the rules of logarithms we can re-write as [(1/50)log a] / [(1/2)log a] Cancel down by 'log a' hence (1/50) / ( 1/2) 1/50 X 2/1 = 2/50 = 1/25 the answer!!!!!!
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
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Three basic log rules when dealing with logarithms are multiplication inside the log can be turned into addition outside the log, division inside the log can be turned into subtraction outside the log, an exponent on everything inside a log can be moved out front as a multiplier vice versa.
k=log4 91.8 4^k=91.8 -- b/c of log rules-- log 4^k=log 91.8 -- b/c of log rules-- k*log 4=log91.8 --> divide by log 4 k=log 91.8/log 4 k= 3.260
this is the question (log (base2) (x))^2- 12(log (base 2) (x)) + 32 = 0, I don't get this bit (log (base2) (x))^2, note the whole log is squared
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19, of course. Review your logarithmic rules.
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you get kamara every 30 days that you log on. If you misbehave or disobey the rules you get kamara taken away..
== == Since each of the log rules was developed using different methods with different assumptions, it is logical that they will not always result in the same volumes for given size logs. Table 1 shows a comparison of the three log rules for 16-foot logs ranging in diameter from 6 to 40 inches. Compared to the International Rule, both the Scribner and Doyle Rules underscale logs of smaller diameters. For example, a 12-inch-diameter log contains 95 board feet on the International scale, 80 board feet on the Scribner scale, and 64 board feet on the Doyle scale. Overall, the Doyle Rule will result in lower log volumes than the International Rule, up to a log diameter of 30 inches. Since nearly all logs in Virginia are below 30 inches in diameter, for all practical purposes the Doyle Rule will underestimate the actual board footage. If you are selling stumpage or logs, it is important to recognize the differences in volume associated with the different log rules. Since stumpage or log prices are based on the timber or log volume, you will receive substantially more income with the rules that scale your sizes higher. For example, take a log of 16 inches diameter x 16 feet in length and a value of $100 per thousand board feet. This log would have the following volumes and values based upon the different log rules: {| ! LogDiameter(in.) ! Log Rule ! Volume(board feet) ! Value ($) | 16 International 180 18.00 16 Scribner Decimal C 160 16.00 16 Doyle 144 14.40 |}
It means you have lost your acces to the RPG becasue you have broken the rules.
250(x) = 400,000 Use logs to base '10' Hence log250^(x) = log400,000 xlog250 = log 400,000 Notirce how the power (x) becomes the coefficient. This is oerfectly correct under 'log' rules. x = log 400,000 / log 250 (NOT log(250/400,000). x = 5.60206 / 2.39794 x = 2.33619689..... ~ 2.33619 to 5 d.p.
Some basic logarithm rules appear in related links.