# Help With A simple maths problem

• October 11th 2010, 03:56 AM
dubbeat
Help With A simple maths problem
Hi,

First off the bat I need to say that I'm so bad at maths I'm not even sure what category this math problem belongs in so apologies if I've posted it in the wrong area.

The Problem:

I have this formula that I use to calculate a decibel level.

db_gain=(x * (8 / 5)) - 90;

When Using this formula db_gain value is not known and x value is known.

How can I switch around? I.E db_gain is known but x is not?

Thanks,

dub
• October 11th 2010, 04:53 AM
Quote:

Originally Posted by dubbeat
Hi,

First off the bat I need to say that I'm so bad at maths I'm not even sure what category this math problem belongs in so apologies if I've posted it in the wrong area.

The Problem:

I have this formula that I use to calculate a decibel level.

db_gain=(x * (8 / 5)) - 90;

When Using this formula db_gain value is not known and x value is known.

How can I switch around? I.E db_gain is known but x is not?

Thanks,

dub

The formula has db_gain "in terms of x" so that you can calculate db_gain if x is known.

You want x "in terms of db_gain" so that you can calculate x from db_gain.

This requires moving the constants across to db_gain,
either using rules or knowing why..

$\displaystyle\ db_-gain=x\ \frac{8}{5}-90$

$\displaystyle\ db_-gain+90=x\ \frac{8}{5}+90-90=x\ \frac{8}{5}$

$\displaystyle\ (db_-gain+90)\ \frac{5}{8}=x\ \frac{8}{5}\ \frac{5}{8}=x$
• October 11th 2010, 04:59 AM
dubbeat
thanks for that. Showing the steps involved is especially useful