Weird logarithmic equations

1. Feb 21, 2007

Essnov

1. The problem statement, all variables and given/known data
I got this problem in an exam today, and I think I may have the right answer but I don't understand how I'm supposed to know if I do. It's very confusing.

2. Relevant equations
If logb (9) = 1.5 and logb (2) = 0.5 what is logb (5throot(36))?

3. The attempt at a solution
So here it seemed futile to try and find b because it seems to have 2 different values at once. Here's what I did...:

logb 9 + logb 2 + logb 2 = 1.5 + 0.5 + 0.5
logb 36 = 2.5
1/5 logb 36 = 0.5
logb 5throot(36) = 0.5

2. Feb 21, 2007

robb_

I am confused on your notation. What base are we working in? b or ()

3. Feb 21, 2007

Essnov

It is base b.

4. Feb 22, 2007

HallsofIvy

Staff Emeritus
What you've done is completely correct. If $log_b 9= 1.5$ and $log_b= .05$, then $log_b ^5\sqrt{36}= 0.5$. But this is a very peculiar question- you are also correct that there is no value of b satisfying those conditions. In fact, $log_b 2= 0.5$ and $log_b ^5\sqrt{36}= 0.5$ is obviously impossible since $2 \ne ^5\sqrt{36}$!

5. Feb 22, 2007

jing

It does to one decimal place. The question is testing the students' knowledge of the rules of logarithms. The IF in the question does give some leeway.

Essnov is obviously smart enough to realise that log_b(9)=1.5 and log_b(2)=0.5 implies two different values for b, which confused him (her?)

So in trying to keep the question simple (introducing an explanation about decimal places into the question would just misdirect a lot of students) it has placed concerns into Essnovs mind.

Well done Essnov for answering it as it was wanted despite this.

Last edited: Feb 22, 2007
6. Feb 22, 2007

AlephZero

Maths is quite hard enough when it is written down precisely.

Pretending that 2^5 = 36, not 32 isn't maths, it's nonsense.

7. Feb 22, 2007

jing

If log_b(9)=1.5 how do you write b precisely?

8. Feb 22, 2007

cristo

Staff Emeritus
$$log_b9=\frac{ln9}{lnb}=1.5 \Rightarrow lnb=\frac{ln9}{1.5} \Rightarrow b=e^{ln9/1.5}$$

9. Feb 22, 2007

AlephZero

$$b = 81^{1/3}$$ would be another way.

10. Feb 22, 2007

jing

I agree. Now suppose this was about learning to use a mathematical technique in say an engineering context and hence you needed a decimal number so you could apply a measurement. How precise would you give the answer?

11. Feb 22, 2007

cristo

Staff Emeritus
In this case, I would use more than one decimal place to avoid the confusion that has arisen! You say

Now, if I was teaching, I would set the question such that the "smart enough" students are not punished for being intelligent enough to find the question confusing.

But then, I'm not a teacher. However, there are plenty of questions one can ask to enable the student to practice the rules of logarithms, without causing such confusion!

12. Feb 22, 2007

AlephZero

A better version of the question might be

If logb (9) = x and logb (2) = y, what is logb (5throot(36))?

That question requires the same knowledge of log functions to solve it, but skips the arithmetic.

13. Feb 23, 2007

jing

But using algebra adds an extra level of difficulty.

[as an aside in this case I would state:

When logb (9) = x and logb (2) = y, what is logb (5throot(36))?

]

Would you settle for

If logb(9) = 1.51 and logb(2) = 0.48, what is logb(5throot(36))? ?

14. Mar 2, 2007

Essnov

Thanks all for input.

Got my exam results yesterday - scored 100%.