log x = y has a x intercept at....?
537851
1.  The x-intercept is where y=0.  So solve log x=0.
861527
Stop, stop, stop. If it was f(x)=2logx, f(x) cannot be = to (1/log2)*logx, as  1/log2 is not = 2.\nDon’t hurry, put your question correctly with necessary details and don’t save on ()-s in formulas.
808044
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
672055
To get x intercept set x = 0 and solve for y\ny = -2*0 + 9\ny = 9\nx intecept (0,9)\n\nTo get y intercept set y = 0 and solve for x\n0=-2x+9\n-9 = -2x\n9/2 = x\n\ny intecept (9/2, 0)
826204
Do you mean \n\nlog (base n) (x) = 1/(log (base x) (n))\n\nWell let log (base n) (x) = a\nTherefore x = n^a\n\nTake logs (base x)\nSo 1 = log (base x) (n^a)\n= a * log (base x) (n)\nBut a = log (base n) (x)\nTherefore log (base n) (x) * log (base x) (n) = 1\nie log (base n) (x) = 1/(log (base x) (n)) ............. QED
477145
log(2x+1) - log5  = log x\nlog(2x+1) - log x = log 5\nlog((2x+1)/x)  = log 5\n(2x+1)/x  = 5\n2x+1 = 5x\n3x = 1\nx= 1/3
469518
There is an obvious 1-dimensional continuum of solutions to this equation: the line y=x.\nHowever, there are a continuum of more solutions, most of which are nontrivial:\nNote that log(x^y) = y*log(x). Thus log(x)/log(y) = x/y. This implies that y = x^(y/x) and similarly x = y^(x/y). This is the general form of the solution where the obvious solution is the restriction x/y = 1. \nIf we let x/y = 2, we get 2y = y^2, or y(y - 2) = 0 which gives nondegenerate solution point (4, 2) as mentioned previously. Setting x/y = 3 gives us 3y = y^3 or y^2 - 3 = 0, which gives the nontrivial solution point (3*Sqrt[3], Sqrt[3]) and so on.
807860
Yes, though don't forget the y on the second log:\n\nlogb x^2/3 + logb y^3/4\n\nAlso, if you would have had:\nlogb x^2/3 / y^3/4 it would equal logb x^2/3 - logb y^3/4\n\nlog rules are just like exponent rules: multiplication = addition and division = subtraction
495630
your answer is incorrect.\n\nthis is the correct answer:\n\n18x+9y=27 ----->  y= -2x+3; intercept at y= 3 and x= 3/2\n\nhow do i get it?\n\nwhen intercept at y, the x value must be zero, so put 0 into x, you get y= -(2 * 0) + 3 = 3\n\nwhen intercept at x, the y value must be 0, so put 0 into y, you get 0 = -2x + 3 -----> -2x = -3 ----> 2x = 3 ----> x= 3/2\n\ni got trouble also when  i was in school but i keep on practising\n\ngood luck
230547
x intercepts are the values of x that make y=0.  since y is a product, y will be 0 when one of its factors is 0, that is, y = 0 when (8-x)=0 which makes x=8, or when x+3=0, which makes x=-3.\n\ny intercepts have x=0.  if x=0, y = (8-0)(0+3) = 24.
962408
log4+xlogb=xloga\nx(loga-logb)=log4\nx=log4/(loga-logb)
864283
According to the y=mx+b equation, b = y-intercept.\n\n-4 is b, which is the y intercept.  When x=0, y=-4.\n(0,-4)\n\nThe x-intercept is the value of y when x is zero.\n0 = 4x - 4\n4 = 4x\nx=1\nThus when y=0, x=1. (1,0)
158569
Here is a tutorial...you can change any of the parameters that you mentioned and look at the graph as it is made and changed...really cool if you ask me.\nhttp://www.analyzemath.com/logfunction/logfunction.html
983978
Since 1^n = 1 then y=1.  So of course log y = 0.
304476
y = x^2-5x\n0 = x^2-5x\n\nx^2-5x = 0\n\nx(x-5) = 0\n\nx = 0 or x = 5\n\nx-intercepts : (0,0) and (5,0)\n\nANSWER : two
804884
To find the x intercept, replace y (here, g(x)) with 0.\n\n0 = ln(x+1)\n\nSince ln (1) = 0, x+1 = 1 which means x = 0.
750081
log (ab)= log a + log b\nlog (a^p) = p log (a)\nlog (a/b) = log a - log b !