log (x+3) + log (2-x)=1\nand 2^x+1 =5^1-2x\nand also  p = a/1+be^-kt
209170
>> log (x+3) + log (2-x) = 1\n\nlog( a b ) = log( a ) + log ( b ) so\nlog (x + 3) + log (2 - x) = log( (x + 3)(2 - x) ) = 1\n\nlog(e) = 1 if using natural logarithms, log(10) = 1 if using base 10 logarithms\n\nFor the former you have (x + 3)(2 - x) = -(x^2) - x + 6 = e and for the latter you have -(x^2) - x + 6 = 10.  These are both quadratic equations which should be easy to solve.  For natural logarithms you have\n\n-(x^2) - x + 6 = e\nx^2 + x = 6 - e\nx^2 + x + (1/4) = (6 + (1/4) - e) = (25/4) - e\nx^2 + x + (1/4) = (x + (1/2))^2 = (25/4) - e\nx + (1/2) = +/- sqrt( (25/4) - e )\nx = -(1/2) +/- sqrt( (25/4) - e )\n\nAs a final check on suitability of these values, we see that numerically x = 1.379... or x = -2.379... and so both x + 3 and 2 - x are positive for both values of x and so you have no problem taking log( x + 3 ) and log (2 - x) with either value for x.\n\nWith base 10 logarithms, completing the square [i.e., solving the quadratic] would lead to complex roots:\n\n-(x^2) - x + 6 = 10\nx^2 + x = -4\nx^2 + x + (1/4) = -4 + (1/4) = -(15/4)\nx^2 + x + (1/4) = (x + (1/2))^2 = -(15/4)\nx + (1/2) = +/- (1/2)sqrt(15) i\nx = -(1/2) +/- (1/2)sqrt(15) i\n\nSo probably the natural logarithm was intended. :-)\n\n>> 2^x+1 =5^1-2x\n\nIf you mean (2^x) + 1 = (5^1) - 2x then you have\n(2^x) + 1 = 5 - 2x\n(2^x) = 4 - 2x\nand "by inspection" x = 1 is a solution.\n\nIf you graph both y = 2^x and y = 4 - 2x you find only one point of intersection; alternatively, the derivative of 2^x - 4 + 2x is strictly positive, so the function is strictly increasing and continuous and so has a single zero.\n\nEither way x = 1 is the only real solution to\n(2^x) + 1 = (5^1) - 2x\n\nHowever, if you meant\n2^(x + 1) = 5^(1 - 2x)\nthen you get (x + 1) log(2) = (1 - 2x)log(5)\n(log(2) x) + log(2) = log(5) - (2 log(5) x)\n(2 log(5) + log(2)) x = log(5) - log(2)\n(log(5 * 5 * 2)) x = log(5 / 2)\nx log(50) = log(5/2)\nx = log(5/2) / log(50)\n\n>> p = a/1+be^-kt \n\nIf you meant p = a / (1 + b(e^(-kt)))\nthen with all log's to base e you have\n\na = p (1 + b(e^(-kt)))\nb = ((a / p) - 1) (e^(kt))\ne = inverse log of ( log(b) - log((a / p) - 1) ) / (kt)\nk = - (1/t) log(((a / p) - 1) / b)\np = a / (1 + b(e^(-kt)))\nt = - (1/k) log(((a / p) - 1) / b)
503501
(log x)^3 = log (x^9)\n(log x)^3 = 9 log x\n(log x)^2 = 9\nlog x = +/- 3\nx = 10^3 = 1000  or x = 10^-3 = .001
230456
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
222183
Well, it's hard to tell from the way you wrote\nthis if the right hand side is:\n\n3^T - 1\n\nor if it is:\n\n3^(T-1)\n\nIf it's the first, then I'm stuck too, although,\nnumerically, the answer would be approximately 4.437915.\n\nHowever, if the equation is really:\n\n6*2^T = 3^(T-1)\n\nthen you can solve it using logs.\n\n\nTake the log of both sides:\n\nlog(6*2^t) = log(3^(T-1))\n\nlog(6) + T*log(2) = (T-1) * log(3)\n\nDistribute the (T-1):\n\nlog(6) + T*log(2) = T*log(3) - log(3)\n\nMove the T's to one side and constants to the other:\n\nlog(6)+log(3) = T [ log(3)-log(2) ]\n\nDivide the factors out to get T by itself:\n\n[log(6)+log(3) ] / [ log(3)-log(2) ] = T\n\nPlug this into a calculator, and it appears that\nT is approximately equal to 7.1285.
54147
For any AP,\n \nNth term = A + (n-1)d\nwhere A is first term, d is common difference.\n\n A = log (a)\n\n d = difference between any two consecutive terms\n   =  log(ab) - log(a)\n   =  log(ab/a)     ...... property of logarithm\n   =  log b\n\nHence, nth term is :\n\n T(n) = log(a) + (n-1)log(b)\n\nThat's your answer. I hope this solves your problem.
189881
Is it 2ln(x) + 1 = 3 or 2ln(x+1) = 3\nIf it is 2ln(x) + 1 = 3, then\nsubtract 1 from each side\n2ln(x) = 2  Divide both sides by 2\nln(x) = 1\nPut these in the exponent of e\ne^[ln(x)] = e^1\nx = e^1, so x = e.\n\nIf it is 2ln(x+1) = 3, then divide by 2 on each side\nln(x+1) = 3/2\nPut these in the exponent of e.\ne^[ln(x+1)] = e^[3/2]\nx+1 = e^[3/2]  Subtract 1 from each side\nx = e^[3/2] - 1
269949
Lets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
63615
log_4 3 + 3log_4 2\n\nLet me show you some properties of log\n\nYou know that if b^x = y, then log_b y = x.\nLet pq = x\nm = b^p\nn = b^q\n\nMultiplying these we will get\nmn = (b^p)(b^q)\nmn = b^(p + q)\nConverting to log we will get\nlog_b mn = p + q\n\nSolving for p and q (using the definition), we get\np = log_b m\nq = log_b n\nSubstitute these to the previous equation:\nlog_b mn = p + q\nlog_b mn = log_b m + log_b n\nOr\nlog_b m + log_b n = log_b mn\n\nMeaning if you add logarithms with the same base you can convert them into a single log and multiply the numbers.\n\nWhat if we multiply log, like 5log_b m\nMultiplying it by n means adding the number n times, and since they have the same base you will multiply the inside numbers n times(which also means raising that number to n).In symbols:\nn log_b x = log_b x^n\n\nWe can use these properties for your problem:\n\nlog_4 3 + 3log_4 2\n\nConvert the second log using one of our properties:\n3log_4 2 = log_4 2^3 = log_4 8\n\nSo,\nlog_4 3 + log_4 8\nUse the other property:\n\nlog_4 3 + log_4 8 = log_4 (3*8) = log_4 24\n\nYou cannot simplify it anymore because 24 is not a power of 4. The condensed form is log_4 24.\n\nHope my derivation helps!
108228
a sum of logs is the log of the product, so\nlog4 (x-3) + log4 (x+3)=2\nlog4 [(x-3)(x+3)] = 2...........then by definition of log\n(x-3)(x+3)=4^2\nx^2-9=16\nx^2=25\nx= ±5.........but we can't use -5 because that would make x-3=-8 and log of a negative is not defined for real numbers,\n\nso x=5
643628
I think you need to rephrase this question.  I'm going to solve this problem:\n\n6^(2x)>5^(x+2)\n\nFirst, logarithm both sides and simplify.\n\n2x*ln6>(x+2)*ln5\n\nln5 and ln6 are both positive, so you don't have to worry about changing your sign anytime in the future.  Now just get all of the x's on one side and factor.\n\nx*(2ln6-ln5)>2ln5\n\n2ln6-ln5>0, so no worries in solving for x.\n\nx>(2ln5)/(2ln6-ln5)\n\nor \n\nx>1.631
454920
take the log\nso log b= log (1+(r/n))^n*t\nlog b = nt log (1+(r/n))\n{(log b)} / {log (1+(r/n))} = nt\n[{(log b)} / {log (1+(r/n))}] / n = t\ni hope those parenthesis dont mess u up lol\ni just learned this in school
494377
log 6^(2x)= log 9037984\n\n2x*log 6= log 9037984\n\n2x= log 9037984/log 6\n\nx= log 9037984/(2*log 6)\n\nx= log 9037984/log 6²\n\nx= log 9037984 / log 36\n\n₢
403896
yes, it really is easier than you think:\na^(log_a(2/x) +log_a(x))=a^(log_a(2) -log_a(x) +log_a(x))\n  =a^(log_a(2))\n  =2\nnote that i used the property log(a/b)= log(a)- log(b) and a^log_a(x)=x
370896
2 ^ (y+1) = 3 * 5^Y\ntake log, or ln\nln 2 ^ (y+1) = ln 3 * 5^Y\n(y+1)ln(2) = ln(3) + yln(5)\ny(ln(2)-ln(5)) = ln(3) -ln(2)\ny= (  ln(3) -ln(2) ) / (ln(2)-ln(5))\n=0.405/-0.916\n= -0.442 which is approx the answer in the back of the book.\n  '
422438
Take the log of both sides:\nlog(0.01) = log(2.2^-t/8)\n\nRemember that log(a^b) = b * log(a), so bring the -t/8 in front:\nlog(0.01) = -t/8 * log(2.2)\n\nNow just solve algebraically:\n-t/8 = log(0.01) / log(2.2)\nt = -8 * log(0.01) / log(2.2)\n\nlog(0.01) = -2\nlog(2.2) ≈ 0.342422681\n\nt = (-8 * -2) / 0.0.342422681\nt = 16 / 0.342422681\n\nTherefore:\nt ≈ 46.7258768\n\nAs a double check, you can plug it in your calculator:\n2.2^(-46.7258768/8) = 0.01
404320
log 3^1/2 to the base 3\n=1/2 log 3 to the base 3\n=1/2
643683
this question will require calcs using logarithm.\nhence will take time