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

[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.

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

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, 201015 yr by mississippichem

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.

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 log_b(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, 201015 yr by the tree