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    Use logarithms to solve the equation 3^{2x-1}=0.05, giving your value of x to four decimal places.

    I have done:
    log3^{2x-1}=log0.05
    2x-1log3=log0.05
    2x-log3=log0.05

    However, for the next step I do:
     x = \frac {log0.05}{2log3}

    On the markscheme it is:
     x = \frac {log_{10}0.05}{2log_{10}3}+\frac  {1}{2}
    Where did the half come from?





    And also....
    log_a x = 2(log_a 3 + log_a 2) -1
    Express x in terms of a, giving your answer in a form not involving logarithms.

    I expand the bracket out first (right?), do I start with -1 or 2?
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    You somehow managed to get that (2x-1)\log 3 = 2x-1\log 3. That's like saying (a+b)c = a+bc... simple algebraic error. If you fix this then you get the half.
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    (Original post by Kash:))
    Use logarithms to solve the equation 3^{2x-1}=0.05, giving your value of x to four decimal places.

    I have done:
    log3^{2x-1}=log0.05
    2x-1log3=log0.05
    2x-log3=log0.05

    However, for the next step I do:
     x = \frac {log0.05}{2log3}

    On the markscheme it is:
     x = \frac {log_{10}0.05}{2log_{10}3}+\frac  {1}{2}
    Where did the half come from?
    Second lines wrong, it should be (2x-1)log3.
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    (Original post by Kash:))
    2x<b>-1</b>log3=log0.05
    2x-log3=log0.05


    What did u do to the '1' ?
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    I saw this thread and got COMPLETELY the wrong idea...
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    (Original post by Kash:))
    Use logarithms to solve the equation 3^{2x-1}=0.05, giving your value of x to four decimal places.

    I have done:
    log3^{2x-1}=log0.05
    2x-1log3=log0.05
    2x-log3=log0.05

    However, for the next step I do:
     x = \frac {log0.05}{2log3}

    On the markscheme it is:
     x = \frac {log_{10}0.05}{2log_{10}3}+\frac  {1}{2}
    Where did the half come from?
    If you did your second line correctly you should get something like:
    2x-1=\frac{\log 0.05}{\log 3}
    to solve.
    • Thread Starter
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    (Original post by ColonelMoore)
    I saw this thread and got COMPLETELY the wrong idea...
    I guess your not a maths a level student then.


    Thanks for the quick reply guys :yy:
    • Thread Starter
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    (Original post by Farhan.Hanif93)
    If you did your second line correctly you should get something like:
    2x-1=\frac{\log 0.05}{\log 2}
    to solve.
    you mean over log3, right?
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    (Original post by Kash:))
    you mean over log3, right?
    Sorry typo.
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    (Original post by Kash:))
    And also....
    log_a x = 2(log_a 3 + log_a 2) -1
    Express x in terms of a, giving your answer in a form not involving logarithms.

    I expand the bracket out first (right?), do I start with -1 or 2?
    the 2 first so it becomeslog_a x = 2log_a 3 + 2log_a 2 -1

    log_a x=log_a 3^2 +log_a 2^2 -1
    log_a x=log_a 9 +log_a 4 -1
    • Thread Starter
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    I still don't understand this.

    I have

    (2x-1)=\frac{log0.05}{log3}
    x = \frac {log0.5}{log3} + \frac {1}{2}

    I am missing the 2log3, why's that.

    Sorry if i'm being thick.
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    (Original post by Kash:))
    I still don't understand this.

    I have

    (2x-1)=\frac{log0.05}{log3}
    x = \frac {log0.5}{log3} + \frac {1}{2}

    I am missing the 2log3, why's that.

    Sorry if i'm being thick.
    (2x-1)=\frac{log0.05}{log3} add 1 to both sides then divide by 2 on both parts of other side not just the one
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    (Original post by Kash:))
    I still don't understand this.

    I have

    (2x-1)=\frac{log0.05}{log3}
    x = \frac {log0.5}{log3} + \frac {1}{2}

    I am missing the 2log3, why's that.

    Sorry if i'm being thick.
    Because your dividing everything by 2 - this means the denominator is going to be multiplied by 2
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    (Original post by Kash:))
    I still don't understand this.

    I have

    (2x-1)=\frac{log0.05}{log3}
    x = \frac {log0.5}{log3} + \frac {1}{2}

    I am missing the 2log3, why's that.

    Sorry if i'm being thick.
    It would be something like this:
     x = \frac {1}{2}\left(\frac {log0.5}{log3} + 1\right)
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    (Original post by Kash:))
    And also....
    log_a x = 2(log_a 3 + log_a 2) -1
    Express x in terms of a, giving your answer in a form not involving logarithms.

    I expand the bracket out first (right?), do I start with -1 or 2?
    I would start off by using the law \log _x y + \log _x z = \log _x yz on the brackets then rewrite the -1 as  -\log _a a
    Then rearrange to get rid of the logs.
    i.e. if done correctly you'll have  \log _a x = \log _a \frac{36}{a} then get rid of the logs.
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    Could someone confirm that "-0.8634" is the correct answer when "giving your value of x to four decimal places"...?

    x
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    (Original post by Farhan.Hanif93)
    I would start off by using the law \log _x y + \log _x z = \log _x yz on the brackets then rewrite the -1 as  -\log _a a
    Then rearrange to get rid of the logs.
    i.e. if done correctly you'll have  \log _a x = \log _a \frac{36}{a} then get rid of the logs.
    To be honest, with the knowledge (with all due repsect) to the OP I would leave the -1 and

    As you say, combine logs in bracket
    subtract that from both sides
    apply quotient property of logs
    then anti log both sides
    cross multiply

    I like what you have said but it may be a jump too far based on the OPs knowledge displayed

    To the OP:

    If you have lg(something +/- something)
    Leave that something alone
    I think you are confusing it with the product and quotient property of logs

    Its the 'quantity' logged rather than a combination of logs
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    (Original post by LJbrowning)
    To be honest, with the knowledge (with all due repsect) to the OP I would leave the -1 and

    As you say, combine logs in bracket
    subtract that from both sides
    apply quotient property of logs
    then anti log both sides
    cross multiply

    I like what you have said but it may be a jump too far based on the OPs knowledge displayed

    To the OP:

    If you have lg(something +/- something)
    Leave that something alone
    I think you are confusing it with the product and quotient property of logs

    Its the 'quantity' logged rather than a combination of logs
    Well I mentioned that to avoid the OP making a mistake with the -1 when getting everything back in exponential form. Also your method will require the OP to recall the laws of indices from C1 so all in all will be more complicated if he leaves the -1.
    • Thread Starter
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    (Original post by FoOtYdUdE)
    Could someone confirm that "-0.8634" is the correct answer when "giving your value of x to four decimal places"...?

    x
    the markscheme says yes
 
 
 
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