<p>narrowing it down further, 1.9311 < x < 1.9312.</p>
<p>by numerical methods, that's a very good approximation.</p>
<p>therefore, x ~ 1.9311 by numerical methods</p>
<p>Yeah, that's a much better approximation than my lame Palm pilot graphing program...Might have to trade it in...</p>
<p>unluckycharms, going by your line of thought, i could make them have the same base but don't know where to go from there. by playing with numbers:
5^x + 9^x = 92 =>
3^1.465x + 3^2x = 92
but i don't know where to go from there. but as far as i know, the possible methods are:</p>
<p>graphically, as some suggested
algebraically with interval bisection, like i did
or iterative methods [a good one is the newton-raphson method]</p>
<p>newton's formula gives the best approximation:</p>
<p>x[2] = x[1] - f(x[1])/f'(x[1])</p>
<p>so you can start with -0.5 as x[1] and go on with a step increase of 0.05 and you continue till you get three constant answers. works best if you have a calculator that can do the job faster.</p>
<p>yup, there was this type of a question in the Australasian (Australia and Asia and New Zealand) maths comp I did last year ... it was one of the first questions i think .. the closest and best approximation was given by the correct application of Newton's method! ....</p>
<p>yeah, but that's where the element of patience and perseverance comes in mathematics and you know, the smaller the step increase, the more accurate your answer and that could take up to 30 lines or more!</p>
<p>yah, there's surely a better way .... but thats how i did it in that test ... i got the 3 marks (out of 3) .... wher's this question from anyway?</p>
<p>well i don't know. where's Newbyreborn? [s]he is the OP.</p>
<p>well, by inspection f(1) < x < f(2), it would be very sensible to begin with 1 or even 1.5, so that would save you a few lines, lol.</p>
<p>just fyi jrock...using step increments to make an approximation is euler's method, not that it's that important.</p>
<p>yeah thanks akai. in my textbook, there actually was no name given to that method (which i called interval bisection by numerical methods). thanks.</p>
<p>I got x = 1.93 by just making an iterative formula and using it. I used the formula y(n+1) = (92 - yn)^(lg5/lg9) and the fact that 5^x = y (note that you need to make sure you have the necessary conditions for convergence). It's a faster way than interval bisection I think.</p>
<p>wow. the first 2 answers i saw were wrong(as already stated) b/c obviously those people didn't know log.. moving on. amrik, can you explain your answer in more detail?</p>