Evaluate logarithmic expression without a calculator

[Helenam]

Hi, I’m really struggling to evaluate this type of expression and appear to be going round in circles. I know the answer is 4 but cannot het there. The question os as follows:
Log_6(3)^2 + log_6 9.log_6 12 + log_6 (12)^2
I hope my notation is understandable as I cannot type it in exactly as it appears in my text book. The first and last parts are log base 6 to the power of 2 of 3 and 12 respectively. Thanks.

 

[Helenam]

I took a photo of the problem and attached it to hopefully clarify what I attempted to write.

 

[topsquark]

I took a photo of the problem and attached it to hopefully clarify what I attempted to write. View attachment 33718

You aren't going to be able to get a decimal expression, if you are looking for that. What I'd suggest is that you break 9 and 12 into their prime factors... [imath]9 = 3^2[/imath] and [imath]12 = 2^2 \cdot 3[/imath]. Then use log rules to simplify the expression into terms with [imath]log_6(2)[/imath] and [imath]log_6(3)[/imath].

Once you've done that you've got one more step left. Can you simplify [imath]log_6(3)[/imath]? (Hint: How can you express 3 in terms of 6?)

Can you finish?

-Dan

Addendum: Also, is [imath]log_6^2 (3) = \left ( log_6 (3) \right )^2[/imath] or [imath]log_6 \left ( \log_6 (3) \right )[/imath]? The second, to my knowledge, is the more standard notation, but the problem suggests the first definition.

 

[Dr.Peterson]

Hi, I’m really struggling to evaluate this type of expression and appear to be going round in circles. I know the answer is 4 but cannot het there. The question os as follows:
Log_6(3)^2 + log_6 9.log_6 12 + log_6 (12)^2
I hope my notation is understandable as I cannot type it in exactly as it appears in my text book. The first and last parts are log base 6 to the power of 2 of 3 and 12 respectively. Thanks.

The answer of 4 is correct; but you'll get there by first simplifying in terms of the logs of 2 and 3.

Show us whatever work you can do, and we can suggest what more to do. (Your typed expression is fine, but you can also show us an image of handwritten work.)

 

[JeffM]

Try this

[math]y = \{ \log_6(3) \}^2 + log_6(9) * log_6(12) + \{log_6(12)\}^2.\\ \text {Let } a = log_6(3) \text { and } b = log_6(12).\\ \therefore \log_6(9) = \log_6(3^2) = 2 * log_6(3) = 2a \implies\\ y = a^2 + 2ab + b^2. [/math]
Now what?

 

[Helenam]

You aren't going to be able to get a decimal expression, if you are looking for that. What I'd suggest is that you break 9 and 12 into their prime factors... [imath]9 = 3^2[/imath] and [imath]12 = 2^2 \cdot 3[/imath]. Then use log rules to simplify the expression into terms with [imath]log_6(2)[/imath] and [imath]log_6(3)[/imath].

Once you've done that you've got one more step left. Can you simplify [imath]log_6(3)[/imath]? (Hint: How can you express 3 in terms of 6?)

Can you finish?

-Dan

Addendum: Also, is [imath]log_6^2 (3) = \left ( log_6 (3) \right )^2[/imath] or [imath]log_6 \left ( \log_6 (3) \right )[/imath]? The second, to my knowledge, is the more standard notation, but the problem suggests the first definition.

Thanks for this. The notation means the first definition. I have tried again using the factors as you suggest but I still don’t know what to do when I have log_6(3)^2. I have had another go as per the attached. Instead of writing log each time I have written just l.

 

[Helenam]

You aren't going to be able to get a decimal expression, if you are looking for that. What I'd suggest is that you break 9 and 12 into their prime factors... [imath]9 = 3^2[/imath] and [imath]12 = 2^2 \cdot 3[/imath]. Then use log rules to simplify the expression into terms with [imath]log_6(2)[/imath] and [imath]log_6(3)[/imath].

Once you've done that you've got one more step left. Can you simplify [imath]log_6(3)[/imath]? (Hint: How can you express 3 in terms of 6?)

Can you finish?

-Dan

Addendum: Also, is [imath]log_6^2 (3) = \left ( log_6 (3) \right )^2[/imath] or [imath]log_6 \left ( \log_6 (3) \right )[/imath]? The second, to my knowledge, is the more standard notation, but the problem suggests the first definition.

Thanks for this. The notation means the first definition. I have tried again using the factors as you suggest but I still don’t know what to do when I have log_6(3)^2. I have had another go as per the attached. Instead of writing log each time I have written just l. View attachment 33721

 

[Helenam]

I think I made some errors in the first attempt but the expression keeps getting more complicated. I’m not sure what you mean by expressing 3 in terms of 6 (18/3)?

 

[Dr.Peterson]

Thanks for this. The notation means the first definition. I have tried again using the factors as you suggest but I still don’t know what to do when I have log_6(3)^2. I have had another go as per the attached. Instead of writing log each time I have written just l. View attachment 33721

You're doing fine -- almost. (What I usually do is to write just "log" without the subscript through most of the work, but put the 6 back at the end, where it will matter.)

It may help to do as was suggested and temporarily replace log 3 with x and log 2 with y (or whatever letters you like).

But first, how did the product of two logs become a sum of the same two logs? The rules don't work quite that way. (Keep it as multiplication!)

Also, you dropped the square on the last term.

 

[Helenam]

I realised both of these errors and think I corrected them in my last post? I am struggling with what to do when multiplying log by log. I’ll try the x and y to see if that helps. Thanks again.

 

[Deleted member 4993]

I realised both of these errors and think I corrected them in my last post? I am struggling with what to do when multiplying log by log. I’ll try the x and y to see if that helps. Thanks again.

\(\displaystyle log_a(x) * log_a(y) = log_a[(x)^ {log_a(y)}]\)

 

[Helenam]

I have attached my latest attempt and it comes to approximately 4 using a calculator but I still cannot simplify the expression to get exactly 4?

 

[Dr.Peterson]

I realised both of these errors and think I corrected them in my last post? I am struggling with what to do when multiplying log by log. I’ll try the x and y to see if that helps. Thanks again.

What you mean is good; but you need parentheses to clearly say it:



Now you'll need to distribute. And using two variables will probably help you keep track of it all! You won't be thinking of it as logs at all, which will also help.

 

[Helenam]

I think I finally got it. Thank you all for your help.

 

[Helenam]

I also managed to simplify it using Jeffm’s suggestion. Thanks again.

 

[Dr.Peterson]

I think I finally got it. Thank you all for your help. View attachment 33725

Good. I did essentially the same, but leaving it in variable form longer:

[math]4x^2+8xy+4y^2=4(x^2+2xy+y^2)=4(x+y)^2=4(\log_6 3+\log_6 2)^2=4\log_6(3\cdot2)^2=4\log_6(6)^2=4[/math]
That looks a lot like what you did Jeff's way.