# Thread: Need help making subject for equation

1. ## Need help making subject for equation

Hey everyone,

I am currently facing a algebra question which have stumped me for days.
I try numerously but I was told that the way I done was wrong.
Can you guys lend me a hand?

I need to make
1) �� the subject in the equation for the first question.
2) �� the subject in the equation for the second question.

Both of the question uses the same equation

b = −(10��������10��+ a)

The questions are done seperately.

2. ## Re: Need help making subject for equation

Originally Posted by colters
Hey everyone,

I am currently facing a algebra question which have stumped me for days.
I try numerously but I was told that the way I done was wrong.
Can you guys lend me a hand?

I need to make
1) �� the subject in the equation for the first question.
2) �� the subject in the equation for the second question.

Both of the question uses the same equation

b = −(10��������10��+ a)

The questions are done seperately.
You're going to have to repost this.

-Dan

3. ## Re: Need help making subject for equation

Originally Posted by colters
Hey everyone,

I am currently facing a algebra question which have stumped me for days.
I try numerously but I was told that the way I done was wrong.
Can you guys lend me a hand?

I need to make
1) �� the subject in the equation for the first question.
2) �� the subject in the equation for the second question.

Both of the question uses the same equation

b = −(10��������10��+ a)

The questions are done seperately.
b= -(10 n log_10(d)+ a)= -10 n log_10(d)- a
a+ b= -10 n log_10(d)

To solve for n, since -10log_10(d) is multiplied by it, divide both sides by those:
-(a+ b)/(10 log_10(d))= n

Going back to a+ b= -10 n log_10(d), to solve for d, first solve for log_10(d) by dividing both sides by -10 n to get
-(a+ b)/(10 n)= log_10(d)

and then "reverse" the logarithm by using its inverse function 10 to the power:
10^(-(a+b)/(10 n))= d

4. ## Re: Need help making subject for equation

Originally Posted by HallsofIvy
b= -(10 n log_10(d)+ a)= -10 n log_10(d)- a
a+ b= -10 n log_10(d)

To solve for n, since -10log_10(d) is multiplied by it, divide both sides by those:
-(a+ b)/(10 log_10(d))= n

Going back to a+ b= -10 n log_10(d), to solve for d, first solve for log_10(d) by dividing both sides by -10 n to get
-(a+ b)/(10 n)= log_10(d)

and then "reverse" the logarithm by using its inverse function 10 to the power:
10^(-(a+b)/(10 n))= d
How did you get that to translate?

-Dan

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6. ## Re: Need help making subject for equation

Dear HallsofIvy & topsquark,

You all have really help me out a lot by solving my problem.

Colters

7. ## Re: Need help making subject for equation

Originally Posted by HallsofIvy
b= -(10 n log_10(d)+ a)= -10 n log_10(d)- a
a+ b= -10 n log_10(d)

To solve for n, since -10log_10(d) is multiplied by it, divide both sides by those:
-(a+ b)/(10 log_10(d))= n

Going back to a+ b= -10 n log_10(d), to solve for d, first solve for log_10(d) by dividing both sides by -10 n to get
-(a+ b)/(10 n)= log_10(d)

and then "reverse" the logarithm by using its inverse function 10 to the power:
10^(-(a+b)/(10 n))= d

left side should equal 10d