ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Fri, 19 May 2017 15:38:34 +0200Problems with sage solve.https://ask.sagemath.org/question/37480/problems-with-sage-solve/ So I have this equation:
eq = 20000 * 1.02^x + 200 * (1 - 1.02^x) / (1 - 1.02) == 30000
And sage gives me this solution with solve(eq, x):
[51^x=4/3⋅50^x]
Why doesnt it give me just the x value?
Please help, thanks.Mon, 01 May 2017 17:40:07 +0200https://ask.sagemath.org/question/37480/problems-with-sage-solve/Answer by ndomes for <p>So I have this equation:</p>
<p>eq = 20000 * 1.02^x + 200 * (1 - 1.02^x) / (1 - 1.02) == 30000</p>
<p>And sage gives me this solution with solve(eq, x):</p>
<p>[51^x=4/3⋅50^x]</p>
<p>Why doesnt it give me just the x value?</p>
<p>Please help, thanks.</p>
https://ask.sagemath.org/question/37480/problems-with-sage-solve/?answer=37484#post-id-37484 eq = 20000 * 1.02^x + 200 * (1 - 1.02^x) / (1 - 1.02) == 30000
eq2 = solve(eq,x)[0]
solve(log(eq2).canonicalize_radical(),x)Mon, 01 May 2017 19:18:14 +0200https://ask.sagemath.org/question/37480/problems-with-sage-solve/?answer=37484#post-id-37484Comment by Ross1856 for <pre><code>eq = 20000 * 1.02^x + 200 * (1 - 1.02^x) / (1 - 1.02) == 30000
eq2 = solve(eq,x)[0]
solve(log(eq2).canonicalize_radical(),x)
</code></pre>
https://ask.sagemath.org/question/37480/problems-with-sage-solve/?comment=37630#post-id-37630I don't understand the answer. Could you be more explicit, please? I'm having a similar problem:
sage: f=4.94 * 1.062^x
sage: solve(f(x) == 15, x)
[531^x == 750/247*500^x]Fri, 19 May 2017 15:38:34 +0200https://ask.sagemath.org/question/37480/problems-with-sage-solve/?comment=37630#post-id-37630