Logarithm problem

[chijioke]

Please I need help with this.
If log 3 = 0.4771, log 5 = 0.6990 and log 7 = 0.8451, evaluate \(\displaystyle \log{ 10.5 } \)
without using table.
This is what I did.

[math]\log{ 10.5 } \\ = \log{\frac{105}{10}} \\ = \log{105}- \log{10} \\ = \log{3×5×7} - \log{2×5} \\ = \log{3} + \log{5} + \log{7} - \log{2} + \log{5} \\ = 0.4771 + 0.8451 + 0.6990 - \log{2} + 0.6690[/math]The challenge now is that in the information given, the value of \(\displaystyle \log{2} \) was not given. Though I know from my calculator that the value of \(\displaystyle \log{2} \) = 0.3010.
My question now is should I use the value 0.3010 for \(\displaystyle \log{2}\) given by calculator to fill in place of \(\displaystyle \log{2}\) or is there another way I am expected to manipulate and finish my work without using the value given by calculator?

 

[Steven G]

In the next to the last line it should be - Log 5 (not + log 5)
You should know the exact value of Log 10.

Hint: How would you write Log_ab = c exponentially (that is without logs)?

 

[Harry_the_cat]

Please I need help with this.
If log 3 = 0.4771, log 5 = 0.6990 and log 7 = 0.8451, evaluate \(\displaystyle \log{ 10.5 } \)
without using table.
This is what I did.

[math]\log{ 10.5 } \\ = \log{\frac{105}{10}} \\ = \log{105}- \log{10} \\ = \log{3×5×7} - \log{2×5} \\ = \log{3} + \log{5} + \log{7} - \log{2} + \log{5} \\ = 0.4771 + 0.8451 + 0.6990 - \log{2} + 0.6690[/math]The challenge now is that in the information given, the value of \(\displaystyle \log{2} \) was not given. Though I know from my calculator that the value of \(\displaystyle \log{2} \) = 0.3010.
My question now is should I use the value 0.3010 for \(\displaystyle \log{2}\) given by calculator to fill in place of \(\displaystyle \log{2}\) or is there another way I am expected to manipulate and finish my work without using the value given by calculator?

You need to insert brackets around 3x5x7 and 2x5 on your fourth line.

log 3 x 5 x 7 =35log3
log(3 x 5 x 7) = log105

 

[Harry_the_cat]

Hint:
\(\displaystyle 2 = \frac{10}{5}\)

or, better still, take Steven's advise after line 3 (You should know the exact value of Log 10), ie don't split the 10 into 2x5 in the first place.

 

[chijioke]

In the next to the last line it should be - Log 5 (not + log 5)

Do you mean? \(\displaystyle \log{3} + \log{5} + \log{7} - \log{2} -\log{5}\)
But even at that you will still have
[math]= \log{3} + \log{5} + \log{7} - \log{2} -\log{5} \\ = \log{3} + \log{7} - \log{2}\\ = 0.4771 + 0.8451[/math]
How do I handle \(\displaystyle \log{2}\)?
Or do I do what was suggested by Harry_the_cat

Hint:
\(\displaystyle 2 = \frac{10}{5}\)

or, better still, take Steven's advise after line 3 (You should know the exact value of Log 10), ie don't split the 10 into 2x5 in the first place.

That is
[math]= \log{3} + \log{7} - \log{\frac{10}{5}}\\ = \log{3} + \log{7} - \log{10} - \log{5} \\ = \log{3} + \log{7} - 1 - \log{5} \\ = 0.4771 + 0.8451 - 1 - 0.6990 \\[/math]

You should know the exact value of Log 10.

[math]= \log{\frac{105}{10}} \\ = \log{105}- \log{10} \\ = \log{3×5×7} - 1 \\ = \log{3} + \log{5} + \log{7} - \log{2} + \log{5} \\ = 0.4771 + 0.8451 + 0.6990 - 1 \\ = 1.0212[/math]

Hint: How would you write Log_ab = c exponentially (that is without logs)?

It would be [math]\log_{ a }{ b } = c \\ a^c = b[/math]

 

[Steven G]

[math]= \log{\frac{105}{10}} \\ = \log{105}- \log{10} \\ = \log{3×5×7} - 1 \\ = \log{3} + \log{5} + \log{7} - \log{2} + \log{5} \\ = 0.4771 + 0.8451 + 0.6990 - 1 \\ = 1.0212[/math]
It would be [math]\log_{ a }{ b } = c \\ a^c = b[/math]

It should be -log 5

Why would you go from 1 to log 2 + log 5 just to go back to log 10 =1????

 

[lookagain]

\(\displaystyle = \log{\frac{105}{10}} \ \)
\(\displaystyle = \log{105}- \log{10} \ \ \)
\(\displaystyle = \log{3×5×7} - 1 \ \ \ \ \ \ \ \ \ \)This is still wrong. Do not continue writing it this way.
\(\displaystyle = \log{3} + \log{5} + \log{7} - \log{2} + \log{5} \ \ \ \ \ \ \) This is an incorrect step for two reasons.
\(\displaystyle = 0.4771 + 0.8451 + 0.6990 - 1 \\ \)
\(\displaystyle = 1.0212\)

Your third line is wrong because of what Harry_the_cat stated in post #3.

Your fourth line is wrong because of what Steven G stated in post #6.

 

[JeffM]

Your answer of [imath]\log (10.5) \approx1.0212[/imath] is correct.

As all the helpers have commented, your process was sloppy

[math]\log (10.5) = \log \left ( \dfrac{105}{10} \right ) \implies \\ \log (10.5) = \log (105) - \log (10) \implies \\ \log (10.5) = \log (105) - 1 \implies \\ \log (10.5) = \log (3 \times 5 \times 7) - 1 \implies \\ \log (10.5) = \log (3) + \log (5) + \log (7) - 1 \implies \\ \log (10.5) \approx 0.4771 + 0.6990 + 0.8451 - 1 \implies \\ \log (10.5) \approx 1.0212. [/math]
The whole issue of [imath]\log (10) = \log (5) + \log 2)[/imath] was an unnecessary complexity given that [imath]\log (10) = 1.[/imath]

 

[chijioke]

Ok I got it now.
[math]\log{ 10.5 } \\ = \log{\left(\frac{ 105 }{ 10 }\right)} \\ = \log{ 105 } - \log{ 10 } \\ = \log{ 3×5×7 } - 1 \\ = 0.4771 + 0.6990 + 0.8451 - 1 \\ = 1.0212[/math]

OR​

[math]\log{ 10.5 } \\ = \log{(\frac{ 105 }{ 10 })} \\ = \log{ 105 } - \log{ 10 } \\ = \log{ 3×5×7 } - (\log{( 2 × 5) }) \\ = \log{ 3 } + \cancel{\log{ 5 }} + \log{ 7 } - \log{ 2 } - \cancel{\log{ 5 }} \\ = \log{ 3 } + \log{ 7 } - \log{ 2 } \\ = \log{ 3 } + \log{ 7 } - \log{\left(\frac{ 10 }{ 5 }\right)} \\ = \log{ 3 } + \log{ 7 } - \left( \log{ 10 } - \log{ 5 } \right) \\ = \log{ 3 } + \log{ 7 } - \log{ 10 } + \log{ 5 } \\ = \log{ 3 } + \log{ 7 } - 1 + \log{ 5 } \\ = 0.4771 + 0.8451 - 1 + 0.6990 \\ = 1.0212[/math]Thanks to everyone that helped me.

 

[Steven G]

There is no reason to convert log 10 to log2 + log 5 only to convert log 2 to log(10/5) = log 10 - log 5.

Yes, the log 5's cancel out as you noted, but then it comes right back. You are told what log 5 equals and know what log 10 equals. Don't touch either one as you know exactly what they equal.

Once again, log3×5×7= log3x(5x7)= (5x7)log 3 = 35 log 3 /= log 3 + log 5 + log 7.
If you want to take the log of 3x5x7, then you write log(3x5x7).

 

[JeffM]

There is no reason to convert log 10 to log2 + log 5 only to convert log 2 to log(10/5) = log 10 - log 5.

Yes, the log 5's cancel out as you noted, but then it comes right back. You are told what log 5 equals and know what log 10 equals. Don't touch either one as you know exactly what they equal.

Once again, log3×5×7= log3x(5x7)= (5x7)log 3 = 35 log 3 /= log 3 + log 5 + log 7.
If you want to take the log of 3x5x7, then you write log(3x5x7).

Actually, Steven, we do NOT know exactly what [imath]\log_{10}(5)[/imath] equals. One of the things that bothers me about this thread is the failure to distinguish between exact answers and approximate answers.

Obviously, your focus in the quoted post was about efficient computation, which is perfectly reasonable. But I believe this poster would benefit by understanding that using logarithm tables gives poorer approximations than does using a decent hand calculator.

 

[Steven G]

Jeff, I do know that the given value for log₁₀(5) is not exact. I was just trying to get the op to stop writing that nonsense time after time.

 

[Deleted member 4993]

we do NOT know exactly what log⁡10(5)\log_{10}(5)log10(5) equals

We know:

log₁₀(5) is exactly equal to log₁₀(5)

Just like:

\(\displaystyle \pi * \sqrt{2} \ \ is \ \ exactly \ \ equal \ \ to\ \pi * \sqrt{2} \) ...............