log[4000(1.0042)^12t - 10,000(1.0063)^12t= 1?
830625
About -37.21\n\nYou want the two terms to be 10 apart...Upon inspection you know t MUST be negative, since the second term can not be greater than the first term.\n\nSo if the two are equal t = -36.55, so through trial and error I got -37.21
541208
it is a proving isn't it?\nright handside: [logbase10 1000]/logbase10 2\n=[logbase10 10^3]/[(logbase2 2 )/( logbase2 10)]\n=3[logbase10 10]x[logbase2 10]\n=3logbase2 10\n=logbase2 1000 , proved\n\n\n\nif its right do mark it best i'll come to know i have done correct or not.\nsorry here is the right answer\nanswer is 4\ni got it by\nlogbase2 1000=1ogbase10 1000 / logbase100 "x"\n3logbase2 10=31ogbase10 10 / logbase100 "x"\nlogbase2 10=1/logbase100 "x"\nlogbase100 10 / logbase100 2 =1/logbase100 "x"\n.5logbase100 100 / logbase100 2 =1/logbase100 "x"\n2logbase100 2=logbase100 "x"\nlogbase100 2^2=logbase100 "x"\n2^2="x"\n"x"=4
820123
log 10 = 1\nln e^-2 = -2\nlog(sub 11) 1 = 0\n\n... -1
13640
24X^4-10X^3-1062X^2+3190X+1124=0
932114
So you have the formula, and you have the number to plug into it. Are you confused about how to deal with the exponents? \n\nIntensity = 5.4 x 10-10, so\n\nd= 10 * log( 5.4 x 10-10/ 10-12)\nd= 10 * log(5.4 / 10-2 )  \nd= 10 * log(540)\n\nd= 10 * 2.732                        note it is a base 10 log!!\nd= 27.32
729730
u cant split a log like that! ie ln(ab) = lna +lnb\n\nu should use the f'(x) f(x)^n = f(x)^(n+1)/ (n+1)
881778
(1.0442) to the power n(number of years)=1.7, the ratio of the final to the original amounts. \nFrom logarithms, n log (1.0442) = log 1.7. considering the same base (let us say 10).\nlog 1.7= 0.2304489214\nlog 1.0442= 0.017867719\nhence, log1.7/log1.0442= 12.2685657176, say approximately in 12 years.
574441
I get\n10.10110111111...\n\nThe question is essentially asking for the value of e in binary. This is no the same as log_2(e). The answer should only have 0 and 1's.
860995
Do the log of both sides.  The exponent will become multiplied by the log of the base.\n\nlog(5/6) = -t/5760 log 2\n\nDivide both sides by log 2, then multiply both sides by -5760 to get t.
1027143
log 2 x^2+1-log2^x\nlog 2 + log X^ + log 1 - (log 2^ + log X)\n=log 2 + A log X + log 1 - A log 2 - log X\n=log 2 + log 1- log X + A log (X - 2)\n=log (2 x 1) - log x + A log (x -2)\n=log 2/x + A log (x-2)\n=log 2/x + log (x-2)^\n= log (2/x X (x-2)^) = log 2/x(x-2)^
151654
It's multiplication.\n\n(64500)(0.02) = 1290\n\nScientific notation is 1.29 * 10^3
372338
Note: 1.^4.4  = 1 ( 1 to the 4.4 power is what you have written.\n\ne/(1.^4.4) = e/1 = e\n\nlog(e/(1.^4.4)) = log(e/1) = log(e) = 1\n\n(2/3)log(e/1^4.4) = 2/3\n\nIs your question stated correctly?
218601
Attached couple of links\n\nApart from these you can buy the book Primer on Logarithms (ISBN 8173714142)
994994
If you are saying \n\n1.) (log(.78))/(log(2)) = -.35845397... or log(2)(.78)\n2.) (log(.80))/(log(2)) = -.32192809... or log(2)(.80)\n\nbut if you mean\n\n1.) (log(78))/(log(2)) = 6.2854022... or log(2)(78)\n2.) (log(80))/(log(2)) = 6.32192809... or log(2)(80)
921114
Multiply by 1000.  40,000
586612
remember your natural logs\n625^x = 1/125\nxln625 = ln(1/125)\nx = ln(1/125)/ln625 = -3/4
706038
125 = log(x^2/1.5) \n125 = 2/1.5 logx  [using logx^m = m logx]\n(125*1.5)/2 = logx\n93.75 = logx\nx = antilog (93.75) OR x = e^93.75