what is the x and y intercept of  -log base 2 (x+2)+3 ?
82029
it's a simple work: I'll give you a suggestion, but you should try to do it by yourself.\n\ny intercept: you find it by setting x=0 so you get\n\ny= log(base2)(2) + 3\n\nThe logarithm is simple to calculate ---> get an integer answer.\n\n---\n\nx intercept: you must set y=0 this time.\nYou get the logarithmic equation\nlog(base2)(x+2) = 3 ---> to solve for x, let each member be the exponent of a power with base 2:\n\n2^(logbase2(x+2)) = 2^3\n\n---> use definition of logarithm (why did we choose a power with base 2?) and get to an integer solution.\n\nGood work.
887169
You're getting into logarithm.  If X/D = R^Y, then Y = log(X/D), where log is the logarithm in base R.
174070
You need to isolate x by using your rules of logarithms.\n\nI'll get you started:\n\ntake the log of every term in the equation\n\nlog2^x+log3^(x+1)=log5\n\nnow all of your exponents can drop to be multiplication on the logs.\n\nxlog2+(x+1)log3=log5\nnow distribute and refactor putting the x's together\n\nxlog2+xlog3+log3=log5\nx(log2+log3)+log3=log5\n\nnow you have one x to work with and a simple algebraic problem
757898
Remember that with logarithms, exponents within the log become the coefficient outside the log.\n\n1.) logc (sqrt x)\n= logc (x^1/2)\n= (1/2) logc x\n= (1/2) * 1.2\n= 0.6\n\nIf numbers are being multiplied within the log, they can be split up to the sum of the log of the numbers.  It's a similar rule to that in problem 1, as you can imagine:\n\nlog (x^2) = log (x * x) = log x + log x = 2 log x\n\n2.) logc xy\n= logc x +logc y\n= 1.2 + 3.9\n= 5.1
754397
ln(x+3) - ln(x) = ln[(x+3)/3]\n\n2ln(2) = ln(2²) = ln4\n\nSo, if ln[(x+3)/x)] = ln4, then\n\n(x+3)/x = 4, or\n\nx+3 = 4x, or \n\nx = 3x, or x = 1\n\nI used the properties of logs to help me solve this.  I think you tried to make it too hard on yourself (which is very easy to do when dealing with the hated logarithms!).  Also, you brought e into it when you didn't really need to.  Since we had ln(x) = ln(y), essentially, then from there we can conclude that x=y without bringing e into play.  Anyhoo, here's the log properties I used in this solution.  (I use the word log but it applies to ln as well.):\n\nlog(a) - log(b) = log(a/b)\n\nrlog(a) = log(a^r)\n\nHope that helped!\n\n~ ♥ ~
733627
Rewrite the equations as:\ny = -x + 2 and\ny = (x+10)/2 = 1/2*x +5\n\nNow, if you insert x=0 in each equation, you will get a y-value.\nIf you insert y=0 in each equation, you will get an x-value.\nFor each line, you can use those two intercepts to graph the line.
537851
1.  The x-intercept is where y=0.  So solve log x=0.
845468
Y-intercepts :   .5\n\nX-intercept:   1\n\nAssymptote at x = 2
760751
a) i)  (0,3)\nThe y-intercept is when x=0, right?  So just plug in zero for x and get:\ny=(0)^2-4(0)+3\nWe get 3 since everything else is zero.\n\nii) (3,0) and (1,0)\nOkay, we have to find when the equation is zero for the x-intercepts.  So, set the equation equal to zero (y=0)\n0=x^2-4x+3\nYou can factor this, which is finding two things that, when multiplied together, form this equation.  The two factors are\n(x-3) and (x-1)  Our factored form is\n(x-3)(x-1)=0\nOne of them MUST equal zero for the whole thing to be zero since they are multiplied, but either one can equal zero.  So we have two possible x values: 3 and 1.  Those are the x-coordinates, the y coordinates are zero.\n\niii) (2,-1)\nThe vertex is halfway in between the roots, which are 3 and 1.  You can take the average of them:  (3+1)/2=4/2=2.  The x-coordinate of the vertex is 2.  Now plug that in:\n\n(2)^2-4(2)+3 = 4-8+3=-4+3=-1\nThe y-coordinate is -1.\n\n\nb) x=2\nThis is just a vertical line with the same value as the x-coordinate of the vertex (the line of symmetry goes through the vertex).  The line is x=2
147089
y intercept: (0,4)\nx intercept: (3,0)
916994
(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
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
829794
The standard equation for slope-intercept form is y=mx+b. m is the slope and b is the y-coordinate of the vertical (y) intercept\n\n-2x=y+3       Original equation\n\ny=-2x-3        Rewritten in S-I form\n\nm= -2, b= -3\n\nThis means that the graph of the funtion is a line through the y-axis at the point (0, -3) with a slope of (-2)
959079
For the y-intercept, set x to zero and solve:\n\nf(0)=-3(0)^2+6(0)-2\nf(0)=-2\n\nFor the x-intercepts, use the quadratic formula since this equation can't be factored. The x-intercepts come out to be 0.42264973 and 1.5773503
867514
[ 1/4 LOG (16) ] - [ 2 LOG (6) ] + [ LOG (3) ]\n\n[ LOG ((16)^(1/4)) ] - [ LOG ((6)^2) ] + [ LOG (3) ]\n\n[ LOG (2) ] - [ LOG (36) ] + [ LOG (3) ]\n\n...........................the first two LOGS have the same base of 10, \n...........................so when subtracting LOGS we divide.\n\n[ LOG (2/36) ] + [ LOG (3) ]\n\n[ LOG ((2divided by 2) / (36 divided by 2)) ] + [ LOG (3) ]\n\n[ LOG (1/18) ] + [ LOG (3) ] \n\n...........................the above two LOGS have the same base of 10, \n...........................so when adding LOGS we multiply.\n\nLOG ((1/18) times (3))\n\nLOG ((1/18) times (3/1))\n\nLOG (3/18)\n\n..................................then reduce the fraction by dividing by 3/3.\n\nLOG ((3 divided by 3) / (18 divided by 3))\n\nLOG (1/6)   \n\nLOG ((6)^(-1))\n\n..........an exponent can be moved and multiplied to the front of the LOG.\n\n(-1) times [ LOG (6) ]\n\n - LOG (6)\n\n\nTherefore, \n\n[ 1/4 LOG (16) ] - [ 2 LOG (6) ] + [ LOG (3) ] =     - LOG (6)
627488
you can change the base...\nln x = (log2 (x))(ln 2)
958491
i think we can convert the log with subsrcipt 3 to natural log\nlogx (with subscript 3) = lnx/ln3\nso derivative is\n1/(xln3)\nhope this helps