wmc Posted December 6, 2010 Share Posted December 6, 2010 I was trying to figure out how to solve a decibel problem. I have been out of school for a very long time and my math is very rusty. Here is an example of the problem: If an amplifier turns a 5watt signal into a 25watt signal thats a gain of ? The equation is: 10xlog10(25/5)=10xlog10(5)=10x0.7=7db now my question. I can easily solve this problem using the log function of my calculator but solving it without a calculator has evaded me. I find that log10=1 but then were did the 0.7 come from. I have attempted to look up the mantissa but I either cannot find the right one in the chart or I am looking for the wrong thing. Can someone show me how to determine the answer of this problem without a calculator. I would also appreciate information on how to use antilogs as well. I believe the antilog is -1 while the log is 1 am I right thanks WMC Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted December 6, 2010 Share Posted December 6, 2010 [math]\log 10 = 1[/math], but [math]\log_{10}(5) \approx 0.7[/math]. [math]\log_{10}[/math] denotes a base 10 logarithm, not the logarithm of 10. So if you wanted to compute [math]\log_{10}(5)[/math] using a table, you'd need to do: [math]\log_{10}(5) = \log_{10}(10 \times 0.5) = \log_{10}(10) + \log_{10}(0.5) = 1 + \log_{10}(0.5)[/math] Then just look up [math]\log_{10}(0.5)[/math] in your table. Link to comment Share on other sites More sharing options...
wmc Posted December 6, 2010 Author Share Posted December 6, 2010 Thank you for your time. I must be a bit dense but how do we know that log10(5) is generally equal to 0.7? Should the number not be 5 then how do I calculate what the in place of (.7) would be? Thanks Link to comment Share on other sites More sharing options...
mississippichem Posted December 6, 2010 Share Posted December 6, 2010 (edited) If [math] log_{a}b=x [/math] then [math] a^{x}=b [/math] so, in your case... [math] log_{10}(5) \approx 0.7[/math] then [math] 10^{0.7} \approx 5 [/math] Edited December 6, 2010 by mississippichem Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted December 6, 2010 Share Posted December 6, 2010 Thank you for your time. I must be a bit dense but how do we know that log10(5) is generally equal to 0.7? Should the number not be 5 then how do I calculate what the in place of (.7) would be? Thanks You'd either need a table of logarithms or a calculator. Numerically calculating a logarithm by hand is hard. Link to comment Share on other sites More sharing options...
the tree Posted December 7, 2010 Share Posted December 7, 2010 (edited) You'd either need a table of logarithms or a calculator. Numerically calculating a logarithm by hand is hard.It's time consuming more than anything, not to mention a little bit pointless. If you want to calculate log10(5) by hand then recall that logb(x) =ln(x)/ln(b)where ln() is the natural logarithm, and you could always calculate that from the integral by the trapezium rule or something. Edited December 7, 2010 by the tree Link to comment Share on other sites More sharing options...
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