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math

posted by on .

2^x+1 =9

how do i solve for this? thanks!

  • math - ,

    2^x + 1 = 9
    2^x = 9 - 1
    2^x = 8
    then we rewrite 8 as power of 2:
    2^x = 2^3
    finally, since they have the same base (which is 2) we equate their exponents:
    x = 3

    hope this helps~ :)

  • math - ,

    sorry, the plus one is part of the exponent
    2^(x+1) =9

    is it possible to move the 1 to the otherside still?

  • math - ,

    Oh sorry. I thought it's outside. Nope, it can't be moved outside.

    2^(x+1) = 9
    2^(x+1) = 3^2
    then we take the log of both sides:
    (x+1)*(log 2) = 2*(log 3)
    x + 1 = 2(log 3) / (log 2)
    x = [2(log 3) / (log 2)] - 1
    or we can also rewrite the 1 as (log 2)/(log 2), so
    x = [2(log 3) - (log 2)] / (log 2)

    hope this helps~ :)

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