Calculus Help: log⁢(x+4)-log⁢(x+3)=1 Solve for x?
817874
Remember that log x - log y =log (x/y) to rewrite that equation. Then, using base 10, you have the equation 10=(x+4/ x+3). Solve.
157110
It would be nice if you could just flip, but...\n\nYou need to change your y to x and your x to y.\n\nFor example:\n\ny=3log(x+6), would become\n\nx=3log(y+6), solve for y \n\nx/3 = log(y+6)\n\n10^(x/3) = y+6\n\ny = 10^(x+3)-6\n\nUse the same process for the others.
100388
This is due to a property of log which is that :\nlog(ab)=log(a)+log(b)\nRemark that this is true when well defined (for eg you can't write log(-1*-3)=log(-1)+log(-3) )\n\nSo here, you have log(x^2)=log(x)+log(x)\n\nTherefore, there are no diff. except that log(x^2) exists for all x in R&#92;{0} and 2log(x) is defined for all x strictly positive.
726536
log 3x² - log 9x = 2\n\nSo log (3x²/9x) = log 100 (Since log a - log b = log (a/b) and 100 = 10²)\nThus\n3x²/9x\n= x/3 \n= 100\nSo x = 300
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.
962408
log4+xlogb=xloga\nx(loga-logb)=log4\nx=log4/(loga-logb)
808003
Do you know your log identities?  That would help you here a lot!\n\nUse these log identities to solve your problems:\n\n. log(a) + log(b) = log(ab)\n. log(a) - log(b) = log(a/b)\n. c*log(a) = log(a^c)\n\nSo, for problem 1 (I'm assuming the "a" means that "a" is your base) we get the following:\n\nlog a(y+5)+2log a (x+1) =\nloga(y+5) + loga((x+1)²) =\nloga((y+5)(x+1)²)  \n\nProblem 2:\n\nIn(x+5)-In(3)=In(x-3)\nln((x+5)/3) = ln(x-3)\n(x+5) / 3 = x-3\nx + 5 = 3x - 9\n2x = 14\nx = 7\n\nProblem 3:\n\n3^x+5=9^x\n\nIs 3 raised to the power of x only, or to the power of x+5?  It's hard to tell, and it does matter to the solution.\n\n~ ♥ ~
886222
Solve for one variable and plug it into the second equation...\n\nSo, let's solve the first equation for y:\n-x + y = -4\ny = x - 4\n\nNow plug this into the second equation:\n4x - 3(x - 4) = 13\n\nDistribute through the parentheses:\n4x - 3x + 12 = 13\n\nSubtract 12 from both sides:\n4x - 3x = 1\n\nSimplify:\nx = 1\n\nNow plug this back into your original substitution:\ny = x - 4\ny = (1) - 4\ny = -3\n\nSo your answer is x = 1 and y = -3.\n\nYou should double-check your answer....\n-x + y = -4,\n-(1) + (-3) =? -4  \n-4 = -4 <-- check\n\n4x - 3y = 13\n4(1) - 3(-3) =? 13\n4 + 9 = 13 <-- check\n\nSo your answer is:\nx = 1\ny = -3
716399
log x-log(x-3)= -1\nlog(x/(x-3))=-1\nx/(x-3)=10^-1=.1\nx=.1(x-3)=.1x-.3\n.9x=-.3\nx=-1/3
636548
e^(3x)=4\n\n3x[log(e)]=log4\nx=[log4]/[3*log(e)]\nx=[ln4]/3\n\nanother solution is to convert e^(3x)=4 to log form directly...if you do that you'll get:\nln4=3x and then transpose the 3...
470800
logx 1/a = (logx 1) - (logx a)\n= 0 - 3\n= -3
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.
713982
use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
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)]
437222
Rules of logs:\n\n--log (a*b) = log a + log b\n--log (a/b) = log a - log b\n--log (a^b) = b log a\n\nSo log (y^2 / x^3) would break down as\n\nlog (y^2 / x^3) = log y^2 - log x^3\n=2 log y - 3 log x\n\nwhich is answer C in your list.
703263
log a (x)= 3.58 and log a (y)= 4.79,\nloga(y/x)=loga(y)-loga(x)\nloga(y/x)=4.79-3.58\nloga(y/x)=1.21
1035877
The natural logarithm (over the real numbers) is defined only for positive numbers, so \n\nx-4>0\n\nin other words\n\nx>4.