# just started transposing at school and having problems with transposing logs

• January 14th 2012, 11:09 PM
leebranch123
just started transposing at school and having problems with transposing logs
Hi everybody, can someone tell me logs are transposed.

Here is one of the problems:

20log10(?) = 35dB

• January 15th 2012, 05:32 AM
skeeter
Re: just started transposing at school and having problems with transposing logs
Quote:

Originally Posted by leebranch123
Hi everybody, can someone tell me logs are transposed.

Here is one of the problems:

20log10(?) = 35dB

let $x = ?$

$20\log_{10}(x) = 35$

divide both sides by 20 ...

$\log_{10}(x) = \frac{7}{4}$

change from a log equation to an exponential equation ...

$x = 10^{7/4} \approx 56.234$

for more info, go to the link and study the three-part lesson on solving log equations ...

Solving Logarithmic Equations from the Definition
• January 15th 2012, 10:57 AM
HallsofIvy
Re: just started transposing at school and having problems with transposing logs
I dislike the term "transposing" because it makes it sound like you are only moving things around. In fact, to solve an equation, for x, you do the opposite of whatever is done to x. To solve x+ 3= 7, because 3 is added to x, and the opposite of "add 3" is "subtract 3", you subtract x (from both sides, of course): (x+ 3)- x= 7- 3, x= 4. To solve 20x= 35, since x is multiplied by 20, and the opposite of "multiply by 20" is "divide b 20", you divide both sides by 20: 20x/20= 35/20, x= 7/4. To solve 20 log(x)= 35, note that two things are done to x- first the logarithm (base 10) and then multiply by 20.

To solve for x, we do the opposite of each step, in the opposite order. That is, we first divide by 20, just as skeeter said. That gives log(x)= 7/4. Then we do the opposite of "logarithm to base 10" which is the "exponential, base 10" (go back and look at you definition of "logarithm"). That is : $10^{log(x))= 10^{7/4}$, x= 10^{7/4}[/tex].