4^x + 6^x = 9^x

find the value of X.




please help me with this!!!

please
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Question

4^x + 6^x = 9^x

find the value of X.

please help me with this!!!

please

in progress 0
Luna 2 years 2021-07-15T14:48:49+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-15T14:50:12+00:00

    \huge \fbox \pink{Solutíon:}

    Given:

     {4}^{x}  +  {6}^{x}  =  {9}^{x}

    OR

    ( \frac{4}{9}  {)}^{x}  + ( \frac{2}{3}  {)}^{x}  = 1

    Putting  \: (2/3) = y

    we Have

     {y}^{2}  + y - 1 = 0

    OR

    y =  \frac{ - 1≠ \sqrt{5} }{2}

    ⇒( \frac{2}{3}  {)}^{x}  =  \frac{ \sqrt{5}  - 1}{2}

    ⇒x = log_{2/3}(  \frac{ \sqrt{5}  - 1}{2} )

     \\  \\  \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}

    0
    2021-07-15T14:50:21+00:00

    Step-by-step explanation:

    Given:

    4^x + 6^x = 9^x

    To find :

    Find the value of x ?

    Solution:

    Given equation is 4^x + 6^x = 9^x

    On dividing by 9^x both sides then

    =>(4^x / 9^x )+ (6^x /9^x) = 9^x /9^x

    =>(4^x / 9^x )+ (6^x /9^x) = 1

    We know that a^m/b^m = (a/b)^m

    =>(4/9)^x +(6/9)^x = 1

    =>(4/9)^x + (2/3)^x = 1

    =>[(2/3)^2]^x +(2/3)^x = 1

    We know that (a^m)^n = a^mn

    =>(2/3)^2x +(2/3)^x = 1

    Put (2/3)^x = a then

    =>a^2 + a = 1

    =>a^2 +2(a)(1/2) = 1

    On adding (1/2)^2 both sides

    =>a^2+2(a)(1/2)+(1/2)^2 = 1+(1/2)^2

    =>[a+(1/2)]^2 = 1+(1/4)

    =>[a+(1/2)]^2 = (4+1)/4

    =>[a+(1/2)]^2 = 5/4

    =>a +(1/2) = ±√(5/4)

    =>a +(1/2) = ±√5/2

    =>a = ±√5/2 -(1/2)

    =>a = ±(√5-1)/2

    Since x is a positive then the square of a positive number is always a positive number

    =>a = (√5-1)/2

    now ,

    (2/3)^x = (√5-1)/2

    On applying logarithmic form

    We know that a^x = N => log N (a) = x

    Wher (a) is base a

    =>x = log (√5-1)/2 (2/3)

    Here base 2/3)

    And it also can be written as

    =>x = log (√5-1)-log2 (2/3)

    (log (√5-1)-log 2 to the base 2/3)

    Answer:

    answer for the given problem is

    x = log (√5-1)/2 (2/3)

    where 2/3 is a base

    Used formulae:

    • a^m/b^m = (a/b)^m
    • (a^m)^n = a^mn
    • a^x = N => log N (a) = x
    • Where a is a base

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