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.

Re: Need help making subject for equation

Quote:

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

Re: Need help making subject for equation

Quote:

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

Add a to both sides:

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

Re: Need help making subject for equation

Quote:

Originally Posted by

**HallsofIvy** b= -(10 n log_10(d)+ a)= -10 n log_10(d)- a

Add a to both sides:

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

Re: Need help making subject for equation

Dear HallsofIvy & topsquark,

Thank you for taking your time to answer my question.

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

Your Sincerely,

Colters

Re: Need help making subject for equation

Quote:

Originally Posted by

**HallsofIvy** b= -(10 n log_10(d)+ a)= -10 n log_10(d)- a

Add a to both sides:

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