document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1112415>Question 1112415</A>: <I><SPAN STYLE=\"word-break: break-all;\"> for the equation y=7 times 10 to the power of 4-x, use base 10 logs to rearrange this equation to make x the subject\r<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 #727441 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=727441\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> start with:\r<BR>\n" );
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document.write( "y = 7 * 10 ^ (4-x)\r<BR>\n" );
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document.write( "take the log of both sides of this equation to get:\r<BR>\n" );
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document.write( "log(y) = log(7 * 10^(4-x))\r<BR>\n" );
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document.write( "since log(a * b) = log(a) + log(b), this equation becomes:\r<BR>\n" );
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document.write( "log(y) = log(7) + log(10^(4-x))\r<BR>\n" );
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document.write( "since log(a^b) = b*log(a), this equation becomes:\r<BR>\n" );
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document.write( "log(y) = log(7) + (4-x) * log(10)\r<BR>\n" );
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document.write( "since log(10) = 1, this equation becomes:\r<BR>\n" );
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document.write( "log(y) = log(7) + 4 - x\r<BR>\n" );
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document.write( "add x to both sides of the equation and subtract log(y) from both sides of the equation to get:\r<BR>\n" );
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document.write( "x = log(7) + 4 - log(y)\r<BR>\n" );
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document.write( "if you wanted to simplify this as far as it will go, then you would get:\r<BR>\n" );
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document.write( "x = 4.84509804 - log(y)\r<BR>\n" );
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document.write( "x is now the subject of the equation.\r<BR>\n" );
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document.write( "to see if this is correct, take any value of y and place it in the equation.\r<BR>\n" );
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document.write( "any value of y > 0 should do.\r<BR>\n" );
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document.write( "i chose y = 15,000,000.\r<BR>\n" );
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document.write( "in the equation of x = log(7) + 4 - log(y), replace y with 15,000,000 to get:\r<BR>\n" );
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document.write( "x = log(7) + 4 - log(15,000,000).\r<BR>\n" );
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document.write( "do not include the commaas when you use your calculator to solve this.\r<BR>\n" );
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document.write( "this results in x =  -2.330993219\r<BR>\n" );
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document.write( "now go back to the original equation of y = 7 * 10 ^ (4-x) and replace x with -2.330993219 to get:\r<BR>\n" );
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document.write( "y = 7 * 10 ^ (4 - (-2.330993219)).\r<BR>\n" );
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document.write( "this results in y = 15,000,000\r<BR>\n" );
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document.write( "the two equations are equivalent.\r<BR>\n" );
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document.write( "one solves for y.<BR>\n" );
document.write( "the other solves for x.\r<BR>\n" );
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document.write( "these two equations are identical.\r<BR>\n" );
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document.write( "if you graphed both these equations, they would draw the same curve on the graph.\r<BR>\n" );
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document.write( "here's the graph of both equations and the intersection of y = 15,000,000 with x = -2.330993219.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2018/031602.jpg\" alt=\"$$$\" >\r<BR>\n" );
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document.write( "you can see that both equations draw the same graph.\r<BR>\n" );
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document.write( "you can see that the intersection of y = 15,000,000 with the graph is at the point (-2.331,15,000,000).\r<BR>\n" );
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document.write( "-2.331, shown as the x-coordinate of the intersection on the graph, is a rounded version of -2.330993219.\r<BR>\n" );
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document.write( "1.5 * 10^7, shown as the y-coordinate of the intersection on the graph, is equal to 15,000,000 in standard decimal form.\r<BR>\n" );
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