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 July 24th, 2013, 08:31 PM #1 Newbie   Joined: Jul 2013 Posts: 2 Thanks: 0 simultaneous log equations Hello all, I am tutoring a high school student, and we both are struggling with the following set of equations: Solve for x and y 1. $\log_{x+1} y= 2$ 2. $\log_{y+1} x= \frac{1}{4}$ So far I have tried re-writing the bases to something like (x+1)(y+1), inverting the bases and arguments and other similar tricks using the change of base formula. All seem to end up with the same two equations: ${(x+1)^2}= y$ and ${(y+1)^{\frac{1}{4}}}= x$ However the resulting quartic equation doesn't yield (beyond one simple solution), and the cubic (after dividing through) is horrible. Any suggestions? Thank you. Peter
 July 24th, 2013, 09:26 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: simultaneous log equations I find: $(x+1)^2=y$ $x^4=y+1$ leading to: $x^3-x^2-2=0$ which has the real solution: $x=\frac{1}{3}$$1+\sqrt[3]{28+3\sqrt{87}}+\sqrt[3]{28-3\sqrt{87}}$$$ and so: $y=$$1+\frac{1}{3}\(1+\sqrt[3]{28+3\sqrt{87}}+\sqrt[3]{28-3\sqrt{87}}$$\)^2$
 July 25th, 2013, 12:48 AM #3 Newbie   Joined: Jul 2013 Posts: 2 Thanks: 0 Re: simultaneous log equations Thanks MarkFL, I got a similar solution for the cubic, using a symbolic solver. I know that the log equations lead directly to the two polynomial equations, but I am still wondering if there is something missed. It seems strange that the student would be expected to solve a cubic, not having covered it in class. Also it seems impossible to check these solutions by using the original log equations. I can however find no fault in the reasoning you gave. Peter
 July 25th, 2013, 02:47 AM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,312 Thanks: 115 Re: simultaneous log equations Who said they were simultaneous equations? Are you sure that you are not simply supposed to do the first step for two independent equations? I.e. the answers are: 1. y = (x + 1)^2 2. y = x^4 - 1 MarkFL's most excellent solution notwithstanding!

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