Converting equation to a logarithm

[Chicago123]

Hello,

Does anyone know how to express this equation to a logarithm? It is likely relatively simple, but I can't get it.

Y=A*X^b*Z^c

Thanks in advance for any help - Really appreciate it!

 

[DrPhil]

Hello,

Does anyone know how to express this equation to a logarithm? It is likely relatively simple, but I can't get it.

Y=A*X^b*Z^c

Thanks in advance for any help - Really appreciate it!

We need to see your work - what have you tried? Where are you stuck? Your first step should be to read about logarithms in your text or notes.

You need to memorize certain rules about logarithms. Logarithms were invented to make multiplication easier: the first rule is
the logarithm of a product is the sum of logarithms.
So in the olden days when you had to multiply a bunch of big numbers, you would look up the logarithms in a table, add them up, and then use the table to find number corresponding to the sum.

Other rules follow from that first rule. For instance division of numbers corresponds to subtraction of logarithms. Multiplying a number by itself doubles the logarithm, or more generally, the logarithm of a power is the power times the logarithm.

 

[Deleted member 4993]

Hello,

Does anyone know how to express this equation to a logarithm? It is likely relatively simple, but I can't get it.

Y=A*X^b*Z^c

Thanks in advance for any help - Really appreciate it!

Start with LOG(Y) = LOG(A*X^b*Z^c)
Use the properties of logarithm: LOG(A * B) = LOG(A) + LOG(B)............. &............. LOG(A^m) = m * LOG(A)

Please share your work with us.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (e.g. "are these correct?")

 

[Chicago123]

Logarithm problem attempt

Hi - Thank you.

I am a college student working to test out of a class via a comprehensive algebra/statistics exam.

Rearranging the equation Y=AX^bZ^C as a log, I get:

Log Y = LogA + bLogX + CLogZ

Will you please let me know if this looks correct/is there anything else to the problem where it can be simplified further.

Thanks.

 

[DrPhil]

Hi - Thank you.

I am a college student working to test out of a class via a comprehensive algebra/statistics exam.

Rearranging the equation Y=AX^bZ^C as a log, I get:

Log Y = LogA + bLogX + CLogZ

Will you please let me know if this looks correct/is there anything else to the problem where it can be simplified further.

Thanks.

That is correct, and complete.

 

[Chicago123]

Thank You

Thank You!