What is the answer to 2^{log_{3}(x)}=\sqrt[a]{x} ?

The answer to 2^{log_{3}(x)}=\sqrt[a]{x} is x=1 {a 0}

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2^{log_{3}(x)}=\sqrt[a]{x}

x = 1 {a < 0 or a > 0}

Solution steps

Apply log rules

lo g_{3} (x) lo g_{3} (2) = \frac{lo g _{3} ( x )}{a}

Multiply both sides by

a

lo g_{3} (x) lo g_{3} (2) a = \frac{lo g _{3} ( x )}{a} a

Simplify

lo g_{3} (x) lo g_{3} (2) a = lo g_{3} (x)

Move

lo g_{3} (x)

to the left side

lo g_{3} (x) lo g_{3} (2) a - lo g_{3} (x) = 0

Factor

lo g_{3} (x) lo g_{3} (2) a - lo g_{3} (x) : lo g_{3} (x) (a lo g_{3} (2) - 1)

lo g_{3} (x) (a lo g_{3} (2) - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

lo g_{3} (x) = 0

Apply log rules

x = 1

Verify Solutions:

x = 1 {a < 0 or a > 0}

The solution is

x = 1 {a < 0 or a > 0}

lo g_{10} (n) = 3600000

lo g_{3} (9 x - 2) = 2 + 4 lo g_{9} (x)

lo g_{10} (m) = 2003

lo g_{x + 2} (x + 2) = - 2

14 lo g_{1} (x^{2} - x - 6) = 1

What is the answer to 2^{log_{3}(x)}=\sqrt[a]{x} ?
The answer to 2^{log_{3}(x)}=\sqrt[a]{x} is x=1 {a<0\lor a>0}