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Simplify the right side of the equation using the exponent rule a^m * aⁿ = a^m+n.

Mathematics

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Answer

64

Step-by-step solution

7 steps

  1. Simplify the right side of the equation using the exponent rule am⋅an=am+na^m \cdot a^n = a^{m+n}.

    xx=2x⋅2320x^x = 2^x \cdot 2^{320} xx=2x+320x^x = 2^{x+320}

  2. Divide both sides by 2x2^x.

    xx2x=2x+3202x\frac{x^x}{2^x} = \frac{2^{x+320}}{2^x}

  3. Simplify both sides using the exponent rules ambm=(ab)m\frac{a^m}{b^m} = \left(\frac{a}{b}\right)^m and aman=am−n\frac{a^m}{a^n} = a^{m-n}.

    (x2)x=2x+320−x\left(\frac{x}{2}\right)^x = 2^{x+320-x} (x2)x=2320\left(\frac{x}{2}\right)^x = 2^{320}

  4. To make the base and exponent on the left side equal, raise both sides to the power of 12\frac{1}{2}.

    This uses the rule (ab)c=abc(a^b)^c = a^{bc}.

    ((x2)x)12=(2320)12\left(\left(\frac{x}{2}\right)^x\right)^{\frac{1}{2}} = \left(2^{320}\right)^{\frac{1}{2}} (x2)x2=23202\left(\frac{x}{2}\right)^{\frac{x}{2}} = 2^{\frac{320}{2}} (x2)x2=2160\left(\frac{x}{2}\right)^{\frac{x}{2}} = 2^{160}

  5. Express the right side in the form kkk^k.

    We need to find a number kk such that kk=2160k^k = 2^{160}.
    We can rewrite 21602^{160} as (2a)b(2^a)^b where a⋅b=160a \cdot b = 160. We want b=2ab = 2^a.
    Let's try a=5a=5. Then 2a=25=322^a = 2^5 = 32.
    If a=5a=5, then b=1605=32b = \frac{160}{5} = 32.
    So, 2160=(25)32=32322^{160} = (2^5)^{32} = 32^{32}.
    The equation becomes:

    (x2)x2=3232\left(\frac{x}{2}\right)^{\frac{x}{2}} = 32^{32}

  6. Equate the bases (and exponents) since they are in the form AA=BBA^A = B^B.

    x2=32\frac{x}{2} = 32

  7. Solve for xx.

    x=32×2x = 32 \times 2 x=64x = 64

    The final answer is 64.

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