Convert log equation in java

Question

i have this equation and im having trouble converting it into java code. The main problem is the "^" operator. I'm not sure how to take a power of an expression.

Here is the equation:

Very quick clarification, its log(5) +1 in the beginning.

Hope someone can help.

Here is the first part of the equation i wrote:

        double equationPart1 =  (((-0.5*Math.log(5)) +1)/2);
        double equationPart2 = 0;//the last part
        double equationPart3 = ((level * Math.log(5)) - (0.5 * Math.log(5)) + equationPart2);

Ever thought of using `pow` from the math library? – Manav Dubey Apr 27 '17 at 00:16 — Manav Dubey, Apr 27 '17 at 00:16
oh can pow take in expressions? – TheQ Apr 27 '17 at 00:21 — TheQ, Apr 27 '17 at 00:21

Devendra Lattu · Accepted Answer · 2017-04-27T00:26:57.813

2

Thumb rule: try to make your equations as readable as possible.

double A = -0.5 * Math.log10(5) + 1;
double B = level * Math.log10(5);
double C = 0.5 * Math.log10(5);
double D = 0.5 * Math.log10(5) - 1;
double E = Math.pow(D/2, 2);

double club = (A/2) + (B - C  + E);    // OR double club = (A/2) - (B - C  + E);
double result = Math.pow(10.0, club);

edited Apr 27 '17 at 00:26

answered Apr 27 '17 at 00:21

Devendra Lattu

2,732
2
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quick question, whats the difference between, Math.log and Math.log10? – TheQ Apr 27 '17 at 00:22
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quick answer - what are javadocs for? – Scary Wombat Apr 27 '17 at 00:23
Lol... `Math.log`: Returns the natural logarithm (base e) of a double value. Whereas later one is for base 10 – Devendra Lattu Apr 27 '17 at 00:24
well i upvoted it to 0. and yeah sounds good. will accept it – TheQ Apr 27 '17 at 00:28
1

Why stop at equations? All code should be written to be as readable as possible. Good answer, though. – duffymo Apr 27 '17 at 09:06

Convert log equation in java

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