Hello everyone,

there should be a relatively easy solution to this but I can't get the correct answer.

\( 20\log_1_0(x) = 20dB\)

How do I find x?

I try to rearrange (taking the exponential of both sides but that does not work). Am I suppost to change the RHS (20dB)?

 

Hello everyone,

there should be a relatively easy solution to this but I can't get the correct answer.

\( 20\log_1_0(x) = 20dB\)

How do I find x?

I try to rearrange (taking the exponential of both sides but that does not work). Am I suppost to change the RHS (20dB)?

how about divide by 20 first?

 

First step

\( \frac{20dB}{20} = 1 \)

Second step

\(10^1 = 10 \)

So the x = 10

for 25dB

\( \frac{25dB}{20} = 1.25 \)

\(10^{1.25} = 17.7827941 \)

 

Wiki covers it well

The part where people get caught up is when to use 10 log or 20 log for dB. When measuring voltage or current gain, use 20 log, when measuring power gain, use 10 log.

It is also good to remember the conversion for dB/Octave <-> dB/Decade (left as an exercise for student)

 

Brilliant, thank you Jony130 and thatoneguy !!