document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1028170>Question 1028170</A>: <I><SPAN STYLE=\"word-break: break-all;\"> We were asked to solve logarithmic inequations, and although I could solve it, I stopped at a dead end, in which I couldn't find a ''good'' answer..\r<BR>\n" );
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document.write( "i) 2*ln(x+2)-1 > 0 \r<BR>\n" );
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document.write( "ii) log(3-2x) > 1 + log(x+5)<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #643346 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=643346\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> i would solve for the equality and then determine the inequality.\r<BR>\n" );
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document.write( "the first equation is 2 * ln(x+2) - 1 > 0\r<BR>\n" );
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document.write( "set that equal to 0 to get 2 * ln(x+2) - 1 = 0\r<BR>\n" );
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document.write( "add 1 to both sides of the equation to get 2 * ln(x+2) = 1\r<BR>\n" );
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document.write( "divide both sides of the equation by 2 to get ln(x+2) = 1/2\r<BR>\n" );
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document.write( "this is true if and only if e^(1/2) = x + 2\r<BR>\n" );
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document.write( "solve for x to get x = e^(1/2) - 2 which results in x = .3512787293\r<BR>\n" );
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document.write( "when x is greater than this, you will find that the equation is positive.\r<BR>\n" );
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document.write( "when x is less than this, you will find that the equation is negative.\r<BR>\n" );
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document.write( "since it's only equal to 0 at one point, this means that all values of x less than -.3512787293 will give you a negative result and all values of x greater than -.3512787293 will give you a positive result.\r<BR>\n" );
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document.write( "therefore, the solution is that 2*ln(x+2)-1 > 0 when x > .3512787293\r<BR>\n" );
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document.write( "here's the graph of the equation.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28600%2C600%2C-5%2C5%2C-20%2C20%2C2%2Aln%28x%2B2%29-1%29 ALIGN=MIDDLE  ALT=\"graph%28600%2C600%2C-5%2C5%2C-20%2C20%2C2%2Aln%28x%2B2%29-1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "your second equation is log(3-2x) > 1 + log(x+5)\r<BR>\n" );
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document.write( "subtract log(x+5) from both sides of the equation to get:\r<BR>\n" );
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document.write( "log(3-2x) - log(x+5) > 1\r<BR>\n" );
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document.write( "since log(a) - log(b) is equal to log(a/b), your equation becomes:\r<BR>\n" );
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document.write( "log((3-2x)/(x+5)) > 1\r<BR>\n" );
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document.write( "set the equation equal to 1 to get:\r<BR>\n" );
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document.write( "log((3-2x)/(x+5)) = 1\r<BR>\n" );
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document.write( "this is true if and only if 10^1 = (3-2x) / (x+5)\r<BR>\n" );
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document.write( "simplify to get 10 = (3-2x) / (x+5)\r<BR>\n" );
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document.write( "multiply both sides of this equation by (x+5) to get:\r<BR>\n" );
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document.write( "10 * (x+5) = 3-2x\r<BR>\n" );
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document.write( "simplify to get 10x + 50 = 3 - 2x\r<BR>\n" );
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document.write( "add 2x to both sides of this equation and subtract 50 from both sides of this equation to get:\r<BR>\n" );
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document.write( "12x = -47\r<BR>\n" );
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document.write( "divide both sides of this equation by 12 to get:\r<BR>\n" );
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document.write( "x = -47/12\r<BR>\n" );
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document.write( "if you replace x with -47/12 in the original equation, you will see that the equation is true.\r<BR>\n" );
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document.write( "your original equation is log(3-2x) > 1 + log(x+5)\r<BR>\n" );
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document.write( "this equation will be valid as long as 3-2x is > -0 and as long as (x+5) is > 0\r<BR>\n" );
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document.write( "otherwise the equation becomes invalid because you can't take the log of a number that is not positive.\r<BR>\n" );
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document.write( "for 3 - 2x to be positive, then solve 3 - 2x > 0<BR>\n" );
document.write( "add 2x to both sides of the equation to get 3 > 2x<BR>\n" );
document.write( "divide both sides of this equation  by 2 to get 3/2 > x<BR>\n" );
document.write( "this means that x has to be < 3/2 or 1.5.<BR>\n" );
document.write( "if x is greater than or equal to 1.5, the equation is invalid.\r<BR>\n" );
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document.write( "likewise x + 5 must be > 0<BR>\n" );
document.write( "start with x + 5 > 0<BR>\n" );
document.write( "subtract 5 from both sides of this equation to get x > -5\r<BR>\n" );
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document.write( "you have x must be greater than -5 and less than 1.5\r<BR>\n" );
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document.write( "you know that the equation is equal to 0 when x = -47/12\r<BR>\n" );
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document.write( "when x = 0, the equation of log(3-2x) > 1 + log(x+5) becomes:\r<BR>\n" );
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document.write( "log(3) > 1 + log(5)\r<BR>\n" );
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document.write( "this results in log(3) smaller than 1 + log(5) so the equation is negative when x > -42/12.\r<BR>\n" );
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document.write( "when x = -4, you get log(3-2x) > 1 + log(x+5) becomes:\r<BR>\n" );
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document.write( "log(11) > 1 + log(1) which is true.\r<BR>\n" );
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document.write( "not that x can't be less than or equal to -5 and can't be greater than or equal to 3/2.\r<BR>\n" );
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document.write( "the solution is therefore that:\r<BR>\n" );
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document.write( "log(3-2x) > 1 + log(x+5) when x > -5 and x < 47/12.\r<BR>\n" );
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document.write( "the solution is therefore -5 < x < 47/12\r<BR>\n" );
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document.write( "if you start with log(3-2x) > 1 + log(x+5) and subtract the expression on the right side of the equation from both sides of the equation you will get:\r<BR>\n" );
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document.write( "log(3-2x) - log(x+5) - 1 > 0\r<BR>\n" );
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document.write( "graph the equation of y = log(3-2x) - log(x+5) - 1 and look at the graph.\r<BR>\n" );
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document.write( "the graph is shown below:\r<BR>\n" );
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document.write( "best i can do for now because i ran out of time.\r<BR>\n" );
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document.write( "if you have further questions, send me an email.\r<BR>\n" );
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document.write( "hopefully this will help.\r<BR>\n" );
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