Solve logarithm by hand

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darfie21

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How do I solve these by hand ? Im able to solve by using logarithms.

8^x=60
3^x=40

using logs I did this x=log 60/log 8 and x=log 40/log 3.

I not sure how to do it by hand.
 
darfie21 said:
How do I solve these by hand ? Im able to solve by using logarithms.

8^x=60
3^x=40

using logs I did this x=log 60/log 8 and x=log 40/log 3.

I not sure how to do it by hand.
Click to expand...
8^x=60 .............................(1)
3^x=40 .............................(2)

Dividing (1) by (2)

(8/3)^x = 3/2

taking 'log' on both sides of '='

x * log(8/3) = log(3/2) ...................continue

Now use the definition of 'log' and continue....
 
These are two different problems 1 and 2 not a system of equations. I'm looking for the
to solve for x by hand for 8^x=60 and
solve for x 3^x=40. Sorry if I wasnt clear.
 
Please tell us exactly what the instructions say and what course you are taking.
 
The course is Algebra II.

The question says with no calculator estimate the value of x.

a) 8^x=60
b)3^x=40
 
Can you please state the entire problem in your 1st post in the future.

8^1 = 8
8^2 =64.
8^x = 60
It seems that x will be between 1 and 2. Is that clear? Since 60 is much closer to 64 then 60 is to 8 then I would suspect that x is closer to 2 than 1. I would estimate x=1.9

You try the 2nd one.

For the record, you said that you tried by hand. Well if 8^x = 60 you can easily arrive at x =log(60)/log(8) by hand so I am not clear what you are saying.
 
Thank You for taking the time to answer. I arrived at x =log(60)/log(8) which requires a calculator to answer. The questions says to
answer without calculator.

Question 2 is appox 3.35.
 
If you don't want to use a calculator, the best you can do is an approximate value. We know that 8^2= 64 so 2 is a little too large. Try 1.5. 8^1.5= 16 times sqrt(2), about 16*1.414=22.624. That's much to small so try 1.75= 1 and 3/4. Continue to zero in on the answer. You will be repeatedly taking square roots which can be done without a calculator (though I confess I did use a calculator).

When they say "answer without a calculator" perhaps they mean for you to give "log(60)/log(8)" as the answer, not a numerical answer.
 
darfie21 said:
Thank You for taking the time to answer. I arrived at x =log(60)/log(8) which requires a calculator to answer. The questions says to
answer without calculator.
Click to expand...
No, I disagree with you. The answer is x =log(60)/log(8) and you got this result without a calculator. A calculator would give you an approximate answer but you arrived at the exact answer without a calculator. Why do you think that the answer to 8^x = 60 is not x =log(60)/log(8)? If you do think that x =log(60)/log(8) is the correct answer then why do you want to go further?
 
darfie21 said:
The question says with no calculator estimate the value of x.

a) 8^x=60
b)3^x=40
Click to expand...
darfie21 said:
Question 2 is approx 3.35.
Click to expand...
The exact wording is essential; it is also important to tell us the context of the question. What specific topics were taught before this was asked? Have you been shown any methods for estimating logs?

Assuming what you gave here are the exact wording, and the actual answer shown for the second problem, we still have the question, how close an estimate are they asking for? I would expect it to say, "to the nearest hundredth" if they expect that answer. On the other hand, perhaps they gave you some specific values to work with, e.g. log(3) = 0.477 and log(2) = 0.301 (base ten), from which you could find approximate answers:

(b) x = log(40)/log(3) = (2log(2) + log(10))/log(3) = (2*0.301 + 1)/(0.477) = 3.35​

Please tell us anything extra that might help clarify what methods might be expected.
 
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