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    Logarithms problem. Edexcel examboard, C2.

    Find to three significant figures, the value of x for which:

    3^{2x+1} = 105

    I solved it in this way.

    3^{2x+1} = 105



log(3^{2x+1}) = log(105)



2x+1 log(3) = log(105)



2x+1 = \dfrac{log(105)}{log(3)}



x = (\dfrac{log(105)}{log(3)} \div 2) - 1



x = 1.118...

    However, when I put my solved value of x back into the equation

    3^{2x+1} = 105

    I get the answer 35, not 105?

    Please could someone tell me where I am going wrong?
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    You made an error on your fifth line. Think about what happens when you divide through by 2 first.
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    (Original post by Ben121)
    You made an error on your fifth line. Think about what happens when you divide through by 2 first.
    I still don't know what I've done wrong, sorry.
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    ( \frac{log(105)}{log(3)} \div 2 ) - 1 should be ( \frac{log(105)}{log(3)} -1 ) \div 2
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    Subtract 1 then divide by 2, not the other way round.
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    (Original post by Wilko94)
    I still don't know what I've done wrong, sorry.
    Ok - I've realised I need to divide by 2 after I minus the 1.

    Please could you explain to me why?
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    (Original post by jameswhughes)
    Subtract 1 then divide by 2, not the other way round.
    I did just recognise that before you posted, but thank you anyway. Please could you explain to me why it works that way?
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    (Original post by Wilko94)
    Ok - I've realised I need to divide by 2 after I minus the 1.

    Please could you explain to me why?
    You are trying to find x from 2x+1, so if you divide by 2 first you end up with x+0.5, not x or x+1
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    It's simple order of method,
    for example..
    if you had
     2x+1 = 5 you would subtract 1 from both sides before dividing by two, to give you  x = 2 . If you were to divide by two first, you would have an incorrect answer of  x = 1.5 . If you insist on dividing first, you must divide the left hand side completely by 2, leaving  x + \frac{1}{2} = 2.5 followed by the subtraction giving the correct answer of  x = 2
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    (Original post by jameswhughes)
    You are trying to find x from 2x+1, so if you divide by 2 first you end up with x+0.5, not x or x+1
    Thank you very much.
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    I don't understand why you have to go the long way, you could just easily do

    log3(105) = x (use calculator to find value, log is to the base 3) you should have a scientific calculator make the most of it

    Then (x-1) divide by 2.

    Wouldn't that be much easier? Thats how I did C2 Logs.
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    (Original post by Scottishsiv)
    I don't understand why you have to go the long way, you could just easily do

    log3(105) = x (use calculator to find value, log is to the base 3) you should have a scientific calculator make the most of it

    Then (x-1) divide by 2.

    Wouldn't that be much easier? Thats how I did C2 Logs.
    I have a scientific calculator - how do I change my calculator to base 3 instead of base 10?
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    (Original post by Scottishsiv)
    I don't understand why you have to go the long way, you could just easily do

    log3(105) = x (use calculator to find value, log is to the base 3) you should have a scientific calculator make the most of it

    Then (x-1) divide by 2.

    Wouldn't that be much easier? Thats how I did C2 Logs.
    My calculator is quite old, it can only do base 10 or natural logs.
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    Hmm, well, you could use

    Log3(105) = 2x + 1

    And Log3(105) = Log10(105)/log10(3)?

    Edit: Log conversion rule

    When you get the answer, check this.
    Spoiler:
    Show

    From that, I get that 2x + 1 = 4.24,
    2x = 3.24
    x = 1.62.
    Which is approximately 105 if you put it back in


    _Kar.
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    I see, but that makes it just as long-winded as the other way IMO.

    I have another problem I'm stuck on.

    

2^{2x-1} - 2^x - 21 = 0

    My teacher told me to solve it by letting  2^x = y , which gives me  2y^2 - y - 21
    which I then factorised to  (2y-7)(y+3) .

    What do I do now? :confused:
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    (Original post by Wilko94)
    I see, but that makes it just as long-winded as the other way IMO.

    I have another problem I'm stuck on.

    

2^{2x-1} - 2^x - 21 = 0

    My teacher told me to solve it by letting  2^x = y , which gives me  2y^2 - y - 21
    which I then factorised to  (2y-7)(y+3) .

    What do I do now? :confused:
    Solve the quadratic, so you get y=3.5 and y=3, then re-write y as 2^x, then take logs, so x=(log3.5)/(log2). The negative root is invalid as logs can only be taken of positive numbers.
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    (Original post by jameswhughes)
    Solve the quadratic, so you get y=3.5 and y=3, then re-write y as 2^x, then take logs, so x=(log3.5)/(log2). The negative root is invalid as logs can only be taken of positive numbers.
    Thanks - I didn't know about the negative thing. I would give you another +rep if I could.
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    Oops, made a mistake in my previous comment. It should say y=-3. Thanks for the rep though
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    (Original post by Wilko94)
    I have a scientific calculator - how do I change my calculator to base 3 instead of base 10?
    if you have the casio calculator fx-82 or above (if you don't you really should buy one, they are really good, you can practically do anything mathematical on them and they are allowed in the exam) it is the button underneath the on button and input the numbers and you are all set


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    (Original post by jameswhughes)
    Oops, made a mistake in my previous comment. It should say y=-3. Thanks for the rep though
    I saw - I just ignored it.
 
 
 
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