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the legs of a right triangle are equal
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to 9 raised to the^ of X and 6 ra to
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the^ of X and the hypotenuse is equal to
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4 raed to the^ of X find X as usual
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pause the video and try to solve it
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yourself according to the Pythagorean
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theorem the sum of the squares of the
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legs is equal to the square of the
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hypotenuse this means that 9 ra to^ of x
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+ 6 ra^ of x² must be equal to 4 raised
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to the^ of x^ 2 when raising a power to
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a power the exponents are multiplied so
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the equation can be Rewritten as 9 ra^ 2
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x + 6 ra^ 2 x = 4 raed to the^ of 2 x
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for any value of the variable X is
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always strictly greater than Z therefore
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we can divide both sides of the equation
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by 4 raed to the^ of 2x the first term
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can then be Rewritten as 94s raed to
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the^ of 2x and the second term can be
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Rewritten as 64s raised to the^ of 2x on
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the right side after simplifying we get
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1 94s is equal to 3 halves and all of
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this is still raised to the power of 2x
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64s after simplifying by dividing by two
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gives us three halves when raising a
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power to a power the exponents multiply
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so the first term can now be written as
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3 raised to the^ of 2x both terms 3v
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raised to the power of 2x can be
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substituted with a new variable for
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example t as a result we obtain a
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quadratic equation t^ 2 + t - 1 = 0
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using the discriminant we find the roots
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T will be equal to - 1 plus orus theare
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< TK of 5 / 2 since T is strictly
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greater than Z we take only the positive
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root now returning to the original
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variable by the definition of logarithms
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we find 2 x 2x will be equal to the
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logarithm with a base of 3v of the < TK
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of 5 - 1 all ID 2 dividing both sides of
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the equation by two we find X
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alternatively 1 12 can be moved to the
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exponent under the logarithmic
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expression the final answer is that the
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value of x is equal to the log arithm
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with a base of 3es of the < TK of 5 - 1
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all divided 2 the problem is solved if
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you understood the solution leave a like
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