solve this 5 + log base 2 ^3?

Okay so 2^z if z= 5 + log(base2) 3 solve for z 

Need help asap

3 Answers

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  • 2 months ago

    First possibility:

    = 2^(z) → where: z= 5 + Log_2(3) = Log[2](96)

    = 2^{5 + Log_2(3)]

    = 2^(5) * 2^[Log_2(3)]

    = 2^(5) * 2^[Ln(3)/Ln(2)]

    = 2^(5) * {2^[Ln(3)]}^[Ln(2)]

    Second possibility:

    z= 5 + Log[2](3) → recall: Log{a}(x) = Ln(x) / Ln(a) ← where a is the base

    z= 5 + [Ln(3)/Ln(2)]

    z= [5.Ln(2) + Ln(3)]/Ln(2)

    z= [Ln(2⁵) + Ln(3)]/Ln(2)

    z= [Ln(32) + Ln(3)]/Ln(2)

    z= [Ln(32 * 3)]/Ln(2)

    z= Ln(96)/Ln(2)

    z= Log[2](96)

    = 2^(z) → where: z= 5 + Log_2(3) = Log[2](96)

    = 2^[Log[2](96)]

  • ?
    Lv 7
    2 months ago

    z = 5 + log₂ 3

                               log 3

                log₂ 3 = --------- ≈ 1.58

                               log 2

    z = 5 + 1.58 = 6.58  ...............ANS

  • rotchm
    Lv 7
    2 months ago

    2^(5 + L3) = 2^5 * 2^(L3) = 2^5 * 3. 

    Can you justify each of these steps?

    Done!

    [PS: Do you really want to solve for z, or to solve for 2^z ?]. 

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