standard form: (1.6 * 10^a) * (8 * 10^b) = c * 10^4

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gortwell

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Hi,

I was given the following problem to solve

(1.6 * 10a) * (8 * 10b) = c * 104

given that the values of a, b and c are in standard form calculate the value of c

My working ....

1.6 * 8 = 12.8
10a * 10b = 10(a+b)

=> 12.8 * 10(a+b) = c * 104

the LHS is not in standard form, rewriting gives

1.28 * 10(a+b-1) = c * 104

therefore c = 1.28
The answer given with the problem is 128 so I'm confused.
 
Romsek said:
1.28*10(a+b+1)=c * 104 not (a+b-1) as you have written.

12.8 = 1.28 * 10
Click to expand...

Yes, sorry for the typo

1.28*10(a+b+1) = c * 104

but this doesn't alter my original question

I have c = 1.28 but the answer given in the book is 128. I am assuming the answer in the book is wrong since the question stated that

c * 104

is in standard form so 1 <= c < 10
 
gortwell said:
I was given the following problem to solve
(1.6 * 10a) * (8 * 10b) = c * 104
given that the values of a, b and c are in standard form calculate the value of c
Click to expand...
gortwell said:
Yes, sorry for the typo
1.28*10(a+b+1) = c * 104
is in standard form so 1 <= c < 10
Click to expand...
I actually see no typo. I find this thread very strange. Lets tease this apart.
\(\displaystyle 1.6\cdot 10^a=16\cdot 10^{a-1}=2^4\cdot 10^{a-1}\)

\(\displaystyle 8\cdot 10^b=2^3\cdot 10^{b}\)

So \(\displaystyle (1.6\cdot 10^a)(8\cdot 10^b)=2^7\cdot 10^{a+b-1}\)

\(\displaystyle 2^7\cdot 10^{a+b-1}=128\cdot 10^{a+b-1}=1.28\cdot 10^{a+b+1}\)
 
I am not sure I understand this problem. But if you take the log of both sides you get:

a+b-log(c/12.8)=4

Is this in standard firm?

Any anyone concur?
 
Quaid said:
If the phrase "standard form" in this thread denotes "scientific notation", then what is the meaning of the following statement? :?
Click to expand...

Hi Quaid,

This is actually my sons homework, he came up with the same answer as me. When he said 'standard form' I said 'um, uh, um scientific notation?' and he said 'um, uh, um?, standard form is what is written in the question!' I assume they mean scientific notation, so 'c' in the original post will be >=1 and <10

in UK, 'standard form' and 'scientific notation' seem to have become synonymous.
The answer I'm looking for is 'No, your answer is wrong, book is right' maybe with a bit of explanation, or 'Yes, your answer is right'.
Some of the replies so far are heading towards the esoteric for absolutely no reason :lol:
 
Last edited:
Putting my chips on your number

gortwell said:
Yes, sorry for the typo

1.28*10(a+b+1) = c * 104

but this doesn't alter my original question

I have c = 1.28 but the answer given in the book is 128. I am assuming the answer in the book is wrong since the question stated that

c * 104

is in standard form so 1 <= c < 10
Click to expand...

Exactly, how can the equality stand unless c = 1.28 and (a+b+1) = 4. We know the 4 is fixed else c(10)^4 would not be in standard form.

Looking at it another way, we want
\[\frac{{1.28{{(10)}^{(a + b + 1)}}}}{{c{{(10)}^4}}} = 1\]

given that the denominator is in standard forum, it again seems that equality holds only if c = 1.28 and (a+b+1) = 4.

If you have a teacher live I hope you will follow up and let us know if the answer was a typo, or there is something more to the definition of "standard form".
 
Well

guitarguy said:
I am not sure I understand this problem. But if you take the log of both sides you get:

a+b-log(c/12.8)=4

Is this in standard firm?

Any anyone concur?
Click to expand...

Your equation looks right although the form might be more "standard" if you move all the constants to one side of the equation, i.e log(12.8); but, I think we will find that standard form = scientific notation. Also, the object is to find a value for c and I can't see how taking the log of both sides will help, but perhaps you have another idea.
 
Dale10101 said:
If you have a teacher live I hope you will follow up and let us know if the answer was a typo, or there is something more to the definition of "standard form".
Click to expand...

Hi Dale10101,

Thanks for the reply ;)

He's back to school on Monday so I'll let you know what happens
 
gortwell said:
This is actually my sons homework


I assume they mean scientific notation, so 'c' in the original post will be >=1 and <10
Click to expand...


I understand the restrictions on c, if the expression c*10^4 represents scientific notation. But why did you post that the number c itself is written in scientific notation?


Maybe you were thinking that the expressions

1.6 * 10^a

0.8 * 10^b

c * 10^4

are written in standard form, but that's not what you posted.

Cheers :cool:
 
oh-oh

mmm4444bot said:
I understand the restrictions on c, if the expression c*10^4 represents scientific notation. But why did you post that the number c itself is written in scientific notation?


Maybe you were thinking that the expressions

1.6 * 10^a

0.8 * 10^b

c * 10^4

are written in standard form, but that's not what you posted.

Cheers :cool:
Click to expand...

Well, there you go, cut me off at the knees and call me shorty.
 
mmm4444bot said:
I understand the restrictions on c, if the expression c*10^4 represents scientific notation. But why did you post that the number c itself is written in scientific notation?


Maybe you were thinking that the expressions

1.6 * 10^a

0.8 * 10^b

c * 10^4

are written in standard form, but that's not what you posted.

Cheers :cool:
Click to expand...

Hi mmm4444bot,

I absolutely understand what you are saying and totally agree.
Having reread the question, it is better worded than I indicated, but my answer still stands.

Here is the question verbatim...

Given that (1.6 * 10a) * (8 * 10b) = (c * 104)
where all three values are in standard form, find the value of c.

(underlining is mine, where I originally, mistakenly, wrote values a, b and c).

Thanks for all the replies and as I replied to Dale, I'll let you know what his teacher has to say about it.
 
book was wrong

Hi all,

My son got his teacher to check his working and she agrees his working was correct and the answer in the book was wrong.

Thanks for all the input and, as an aside for any students reading this, this is one of the many reasons for not cheating by copying the answers from the back of a text book ;)
 
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