multiplying logs (see question posed below?
824419
log 10 * log 10 = 1 because log (base 10) 10 = 1 \nand therefore 1 * 1 = 1\n\nlog 10 + log 10 = 1 + 1 = 2 \nand log 10 + log 10 = log (10*10) = log 100\nlog 100 = 2 means 10^2 = 100 noting again that log means log (base 10) meaning it is the power to which the base is raised to get its argument\n\n1000 = 10^3 so log 1000 = 3\n\nNow using the rules stated\n[ie  log a - log b = log (a/b)\nlog a + log b = log ab\nlog a^b = bloga]\n\nlog 3 + 3 log y = log3 + log (y^5) (since bloga = log a^b)\n= log(3 * y^5) (since log a + log b = log ab)\n= log (3y^5)\n\nThe log of a product = the SUM of the individual logarithms!\n\nlog(ab) = loga PLUS logb defintely NOT loga TIMES log b
725458
Hello dear Sherri \n\n● 4^(2x+1) = 2^(3x+6)\n-Take ' log ' from each sides\n\n● log [4 ^ (2x+1)] = log [2 ^ (3x+6) ]\n● (2x+1) log 4 = (3x+6) log 2\n   { as you know you can get log 4 =  2 log 2 }...\n● 2(2x+1) (log 2) = 3x+6 (log 2)\n   { cancel " log 2 " } ...\n● 4x+2 = 3x+6\n- Now its a simple function \n● 4x - 3x = 6 - 2\n● x = +4\n\nGood Luck Darling ♣
586612
remember your natural logs\n625^x = 1/125\nxln625 = ln(1/125)\nx = ln(1/125)/ln625 = -3/4
658763
Take the logarithm of both sides:\n\nlog a^a = log 2\n\nThen use the rule log (x^y) = y log x\n\nso a log a = log 2\n\nUnfortunately from here we can only solve numerically. Look at the graph of a log a and make approximations.\n\nThough not expressible using elementary functions, the actual answer is defined by the Lambert's W function and reads: log 2 / (LambertW (log 2)), which is roughly 1.5596. More about the W function and what it's used for below.
815091
First of all, we apply the power rule:  a*log(x) = log(x)^a\n\n     2*log(a(x+1)) = log(a(x+1))^2\n\nNext, we know that the sum of two log expressions can be written as a single product: log(a) + log(b) = log(a*b)\n\n     log(a(y+5)) + log(a(x+1))^2 = log[a(y+5)*(a(x+1))^2]\n\nYou may try some algebraic simplification, but I think that these concepts are the ones that are being tested for this problem.
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.
898977
log(2x+1)-(1/3)log(x+1)\n=log(2x+1) - log(  (x+1)^(1/3)  )\n=log( (2x+1) /   (x+1)^(1/3)  )
218601
Attached couple of links\n\nApart from these you can buy the book Primer on Logarithms (ISBN 8173714142)
968058
Type 1.16 log / 4 / 1.0125 log =\nThis is 3 approximately (2.9869...)\n\nTh.
805545
Part A\nlog(base3) 20 = (log 20) / (log 3)\nlog 20 = log (2*10) = log 2 + log 10 = 1.301\nlog 3 = 0.477\n1.301/4.77 = 2.727 --> about 2.73\n\nPart B\nlog(base 9) 100 = (ln 100) / (ln 9)\nln 100 = ln (10^2) = 2 (ln 10)\nln 9 = ln (3^2) = 2 (ln 3)\n(ln 10) / (ln 3) = 2.30 / 1.10 = about 2.10\n\nPart C\n1) log(base a) 6xy^5z^4 = log(base a) 6 + log(base a) x + log(base a) y^5 + log(base a)z^4 = log(base a) 6 + log(base a) x + 5 log(base a) y + 4 log(base a) z\n\n2) log (5a/4b^2) = log 5a - (log 4b^2) = log 5a - (log 4 + log b^2) = log 5a - log 4 - 2 log b\n\nPart D\n(1) 1/2 log a = log (a^(1/2))\nlog (a^(1/2)) - log 2 = log (a^(1/2) / 2)\n\n2) ln 2x + 3 ln x - 3 ln y = ln 2x + ln x^3 - ln y^3 = ln (2x*x^3)/y^3\n= ln (2x^4/y^3)\n\npardon me if you see typos.\nIt's 1:30 in the morning
689473
get the n's on one side\nso\n17959/21558=(1.02/1.03)^n\nor\n.833055014=.990291262^n\nusing logs with change of base\nn=log (.833055014)/log(.990291262)\nor n=18.72
178809
Have you considered some sort of logarithmic compression?  for example, 10*log10(x+1)\nthen 10*log10(1+1) = 3\n10*log10(3+1) = 4\n10*log10(2+1) = 3.7\n10*log10(2.5+1) = 3.?\n10*log10(400 +1) = 26\n10*log10(500 +1) = 27\n\nother types are possible using the natural instead of the base 10 log.   there may be some rounding errors in my numbers, i wasn't using a calculator.\n\nI hope I didn't misunderstand your qustion.
956834
For log(x+8) to be log x +log 8 ,x+8 must be equal to 8x\nWe can write it algebrically as log(x+8)=logx+log8=log(8x)\n=>8x=x+8\n=>7x=8    or,x=8/7
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
911082
logx+log(x+48)=2\n=>log{x(x+48)}=2\n=>log(x^2+48x)=2\n=>x^2+48x=10^2\n=>x^2+48x-100=0\n=>x^2+50x-2x-100=0\n=>x(x+50)-2(x+50)=0\n=>(x+50)(x-2)=0\nhence,\nx=-50 or x=2
118410
...=(logA+logA)-(logB-logB-logB)+1/5logC=(logA^2-logB^3)+1/5logC\n= log(A^3/B^3)+1/5logC =log(A^3/B^3)+logC^1/5=log[(A^3*C^1/5)/B^3]\nNow if you know to transform C^1/5 (i don't really remember now)it becomes even simpler.
643371
First you need to know what logarithms are. (logs)\nAll numbers can be written as a power of ten.  These are logs.  \nUsing a base of ten, \nlog 100 = 2 (because 100= 10 x 10 = 10 to the power of 2)and \nlog 1000 = 3 (because 1000= 10 x 10 x 10 = 10 to the power of 3).  \nNumbers between 100 and 100 will have a log between 2 and 3.  You can find the power by pressing the log button on the calculator.  Often people use natural logs which are powers of e (2.71...). For these press ln\nAnti-logs are the original numbers that the logs represent. Anti-logging is the process of finding the number represented by the log.  These days you would use a calculator and press inv log.  \nFor example log 50 = 1.70\nantilog 1.70 = 50\nwhich is the same as 10 to the power 1.70 = 50.\nIn the olden days, 25 years ago, before calculators Logs were useful when multiplying large numbers. You would find the logs of the numbers and then add the logs, before antilogging the answer to find the result of the multiplication. By subtracting logs you can do division. \n\nNow, just get out the calculator. I haven't heard the term antilog for many years as you would better represent your working using log and then using powers.