log(2x-1)-log(x-3)=1?
815275
Using the quotient rule,\n\nlog(2x-1) - log(x-3)=1\nlog((2x-1)/(x-3)) = 1\n\nThen change to exponential form:\n\n10^1 = (2x-1)/(x-3)\n\nthen solve:\n\n10(x-3)=2x-1\n10x-30=2x-1\n8x=29\nx=29/8
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 ♣
743256
First 4*log[x-2] = log[x-2]^4. \nThen -log[5] = log[1/5]\n\nThe sum of logs is the log of the product of the arguments and so this gives\n\n(1/2)*log[(x-2)^4 * (x+3) / 5]\n\nFinally the (1/2) becomes an exponent in the argument:\n\nlog[((x-2)^4 * (x+3) / 5)^(1/2)]
368201
(1/2)logx = log(x^(1/2)) = log(sqrt(x))\n\nso for exemple 1/2log(9)=log(sqrt(9)) = log(3).\n\nsometimes it's useful, sometimes it's more useful to write the log of a sqrt as 1/2 the log of the argument.
754734
1) The definition of a logarithm says : if a^X = N, then X = log[a]N.\nTurn this around to say : if X = log[a]N, then a^X = N.\nNow you can see what to do if we let X = log[2]64.\nWe get : 2^X = 64 = 2^6.\nTherefore, X = 6, that is, log[2]64 = 6.\n\n2) Same with this one. Let X = log[2]sqrt(2).\nThus, 2^X = sqrt(2) = 2^(1 / 2).\nTherefore, X = 1 / 2.\n\n3) 5^(-3) = 1 / 125\nUse the definition again. If a^X = N, then X = log[a]N.\nHere we have : a = 5, X = -3 and N = 1 / 125.\nThus, -3 = log[5](1 / 125)\n\n4) 10^(0.3010) = 2\nIn the same way, let a = 10, X = 0.3010 and N = 2.\nThus, 0.3010 = log[10]2.\n\n5) t = log[4]7\nDefinition again : 4^t = 7.\n\n6) log[10]7 = 0.845\nDefinition again : 10^(0.845) = 7.\n\n7) Let X = log[64]4\nBy the definiton : 64^X = 4.\nOr, (2^6)^X = 4.\nOr, 2^(6X) = 2^2.\nTherefore, 6X = 2, or, X = 1 / 3.
157812
[x^2/4][2(log(x) - 1) log(x) + 1]
987583
2^(x+1) = 5^(1-2x)\n\nlog 2^(x+1) = log 5^(1-2x)\n\n(x+1) log 2 = (1-2x) log 5\n\n(log 2 + 2 log 5)x = log 5 - log 2\n\nx = (log 5 - log 2)/(log 2 + 2 log 5)
575720
take logbase6 of 135 = x\n\nlog 135/log 6 = x\n2.737 = x\n\nThen substitute this into original to get:\n2^(log135/log6) + 1.3^(log135/log6)-2 \n\nIt should prove that it equals 30, which proves that both statements are ture.  However, I am getting that it is not equal to 30...so I think the proof is impossible,  unless your formula is typed incorrectly.
627488
you can change the base...\nln x = (log2 (x))(ln 2)
790275
log(9)3\n=\nlog(9)9^1/2\n=\n1/2*log(9)9\n=\n1/2\n[as log(m)m=1]
750081
log (ab)= log a + log b\nlog (a^p) = p log (a)\nlog (a/b) = log a - log b !
470727
log_7(7)+log_7(x)=1+log_7(x).
936579
your question seems to be incomplete.\nso far....\nwidth=x\nlength= 2x-1
958049
[Log(11)3 - 7]/0.35 = x\n\nRead: Log base 11 of 3 minus 7 all divided by 0.35 equals x
586612
remember your natural logs\n625^x = 1/125\nxln625 = ln(1/125)\nx = ln(1/125)/ln625 = -3/4
693908
Logs are just equivalent to power statements. \nlog_a (b) = c means a^c = b.\nSo that equation just means x^2 = 2x-3.\nNow thats just a quadratic;\nx^2 - 2x + 3 = 0\nHowever that has no real roots. So there are no such values of x. But hopefully you get the idea.
350639
7^(2x + 1) = 3^(x + 2)\nlog(7)(x + 2) = log(3)(2x + 1)\nIn the second step, you switched the x+2 and 2x+1\nLine 2 should be\nlog(7)(2x + 1) = log(3)(x + 2)\nContinue solving as you did before and you will get the correct answer.\n\nAlso, you wouldn't write ln9 just because it is not in simplest form\nln(9) = ln(3^2) = 2ln(3)\nSimilarly for ln(49) = 2ln(7)