Logarithmic Graph Function True or False Question.?
84854
the exponential (both b>1 and 0<b<1) doesn't intercept the x axis: in fact the x axis is a horizontal asymptote for the function (asymptote = which doesn't touch, by ethimology).\n\nAnyway, if you want an algebrical answer, substitute -1 to x in the function: you get y=b^-1 --> y=1/b\nThe reciprocal of any number is never 0.
41347
The trick to remembering the log rules is that they are the same as the exponent rules since logs are just rewritten exponents.\n\n(x^2)(x^3) = x^(2+3) = x^5\nlog [(a)(b)] = log a + log b\n\n(x^4)/(x^2) = x^(4-2) = x^2\nlog [a/b] = log a - log b\n\n(x^3)^2 = x^[(3)(2)] = x^6\nlog[a^b] = b[log a]\n\nThe rules for log and ln are the same since the only difference is log has base 10 (unless otherwise noted) and ln has a base of e.\nThere are also the two inverse functions with log and ln
239990
There are no values of x for which the function is less than 0.  The function value is between 0 and 1 when x<=0.\n\nPlot a few values to sketch the graph: x=-2,y=1/9; x=-1,y=1/3; x=0,y=1; x=1,y=3; x=2,y=9
985432
well...ln(x) does not cancel e^x....but...\n\ngiven e^x, if we take the natural log of e^x then we have ln(e^x), well this means that we have an exponent...we recall that we can pull the exponent down infront of a logrithm...so we now have xln(e)\n\nah! the ln(e) is 1, we recall that ln's are base e and if we take the logrithm of the base we get 1. so we have x*1=x\n\nso...likewise...lnx^2 is 2ln(x)...and thats all we can do\n\nhope this helps\n\nmatttlocke
736099
plot the coordinates then do the vertical line test.  if the vertical line intersects more than two points on the graph then it is not a function.  each set of coordinates you have listed are functions (i free hand it with no graphing paper or calc) so forgive me if im wrong.  also refer to www.coolmath.com. it is a good site for breaking down math
1052815
3.1 - Given the equation of an exponential function, state its domain and graph it. \n\n3.2 - (a) Given a logarithmic equation, express it as an exponential equation.\n(b) Given an exponential equation, express it as a logarithmic equation. \n\n3.3 - Given the equation of a logarithmic function, state its domain and graph it. \n\n3.4 - (a) State the product, quotient and power properties for logarithmic functions.\n(b) Apply the properties of logarithms to expand and/or simplify expressions involving logarithms. \n\nSolving Exponential and Logarithmic Equations\n- Using the laws of exponents and the properties of logarithms, simplify expressions involving the composition of exponential and logarithmic function of the same base. \n - Given an equation involving logarithmic functions, solve it. \n\n - Given an equation involving exponential functions, solve it. \n\n- Given an applied problem described by an exponential or logarithmic equation, solve the
989452
The parent function is y = 6^x. That function contains the point (0,1). Since f(x) = 6^(x+1) is a translation of 6^x one unit to the left, the point (0,1) maps to (-1,1).  \n\nThe graph of f(x) = 6^(x+1) looks like the graph of 6^x translated one unit to the left.
470658
The above answer is basically correct.  Plug in a few more values, such as (21,1543.5), (22,1694), and so on for a few more values of the graph.
347784
Taking natural logarithm to both sides\nLn(y)= Ln(x) * Ln(x) = (Ln(x))^2\nDifferentiating\n(1/y )dy/dx = (2/x) * (Ln(x))\ndy/dx = (2y/x) * (Ln(x))\nat the point (e,e),\ndy/dx = (2e /e) * Ln(e)\nso the slope = 2
960996
log y = d log (cx)\n\nlogy = d log c + d logx\n\nlet logx = X\nlog y = Y\n\nand d logc = C\n\nthe equation becomes\n\nY=dX+C\n\nthis is of the form y=mx+c which is a straight line
218601
Attached couple of links\n\nApart from these you can buy the book Primer on Logarithms (ISBN 8173714142)
851770
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
364460
log(x) in base a (for a>0) is simply ln(x)/ln(a). So if you are able to calculate natural logarithms, you can calculate logarithms in any base.\nA very simple technique to calculate natural logarithms is to use the identities:\nd/dx(ln(x))=1/x and ln(1) = 0. This implies that to calculate ln(y) for a given y>0, you can use finite approximations to the integral:\n\nintegral form 1 to y of 1/x = ln(y)\n\na very simple approximation is the following: for a given integer (big!) k of your choice, \n\nln(y) = sum for i=0 to k-1 of k/(k-i(y-1))\n\nHope this helps
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.
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.
7970
Log(a^b)= b*Log(a)\nLog(ab) = Log(a)*Log(b)\n\na) Log147= ?\n147= 3*7*7\n=> Log147= Log3 * (Log7)^2\n=> Log147= 2yw\n\nb) Log28= ?\n28= 2*2*7\n=> Log28= (Log2)^2 * Log7\n=> Log28= 2xw
464388
pablodiablo is right, you can't treat an inequality like an equal sign when dividing or multiplying both sides by a negative number...swap the sign when doing so.  if you look at the graph of a logarithmic function, if 0<n<1, then log(n) is a negative number, and since y>x, x/y is less than one.  also, x & y must both be the same sign (positive or negative), otherwise x/y will be negative, and log of a negative number is undefined.