Maths , logarithms

Can someone solve for x please. I don't understand it.

2^x . 7^x = log 14

My answer : 14^x = log 14
= log14÷log 14 = x
X=1
Is this right????

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  1. yes, it is right.

    a^x * b^x= (ab)^x

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    bobpursley
  2. 2^2x+1 - (9)2^x = -4

    Can someone please solve for x ?
    My answer is x= 2 , is this correct?? Please show your steps

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  3. Hi bob
    For the first log problem , it was in a test and my answer was wrong , could it be a marking error?

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  4. is the problem this
    2^(2x+1) -9*2^2x = -4
    or this
    2^(2x) +1 -9*2^2x = -4

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    bobpursley
  5. Sorry it wasn't clear the problem was this 2^(2x+1) -9*2^2x = -4

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  6. yes, it is correct

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    bobpursley
  7. #1 answer is incorrect

    2^x * 6^x = log 14
    14^x = log 14
    now take logs of both sides, and use log rules
    x log14 = log(log14)
    x = log(log14)/log14 = appr .05168

    it is easy to show that x = 1 is not a solution to the question
    since 14^1 ≠ log14

    #2
    2^2x+1 - (9)2^x = -4
    the way you typed it, x = 2 is definitely not a solution
    LS = 4(2)+1-9(4) = -27 ≠ -4

    if you meant:
    2^(2x+1) - (9)2^x = -4 , then
    2(2^(2x) - 9(2^x) + 4 = 0
    2 (2^x)^2 - 9(2^x) + 4 = 0
    let 2^x = y , then we have
    2y^2 - 9y + 4 = 0
    (y - 4)(2y - 1) = 0
    y = 4 or y = 1/2

    then 2^x = 4 -----> x = 2
    or 2^x = 1/2 ------ x = -1

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  8. Thank you Reiny
    Your answer is very clear and shows what I did wrong. Keep up the good work.

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