How do I solve this fractional indices equation 3 5 x + 2 9 1 − x = 27 4 + 3 x 729 ?Solve the equation 3 5 x + 2 9 1 − x = 27 4 + 3 x 729 I thought that the best way of approaching this would be to rewrite everything using 3 as the base of the exponents, hence creating an equivalence which would allow me to equate numerators to numerators, and denominators to denominators. Doing this yields: 3 5 x + 2 3 2 ( 1 − x ) = 3 3 ( 4 + 3 x ) 3 6 Equating the exponents of each numerator: 5 x + 2 = 3 ( 4 + 3 x ) 5 x + 2 = 12 + 9 x − 4 x = 10 x = 10 − 4 = − 5 2 Doing this for the denominator yields a different value of x: 2 ( 1 − x ) = 6 2 − 2 x = 6 − 2 x = 4 x = − 2Why is that I'm obtaining two different values of x? Further to this, the solution in the book states the answer as x = − 3, what am I doing wrong?

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How do I solve this fractional indices equation             3              5        x        +        2                    9              1        −        x              =            27              4        +        3        x              729  ?Solve the equation             3              5        x        +        2                    9              1        −        x              =            27              4        +        3        x              729  I thought that the best way of approaching this would be to rewrite everything using 3 as the base of the exponents, hence creating an equivalence which would allow me to equate numerators to numerators, and denominators to denominators. Doing this yields:            3              5        x        +        2                    3              2        (        1        −        x        )              =            3              3        (        4        +        3        x        )                    3      6      Equating the exponents of each numerator:  5  x  +  2  =  3  (  4  +  3  x  )  5  x  +  2  =  12  +  9  x  −  4  x  =  10  x  =      10          −      4        =  −      5    2  Doing this for the denominator yields a different value of x:  2  (  1  −  x  )  =  6  2  −  2  x  =  6  −  2  x  =  4  x  =  −  2Why is that I&#039;m obtaining two different values of x? Further to this, the solution in the book states the answer as   x  =  −  3, what am I doing wrong?

Nancy Ewing · Accepted Answer

Recall that the property      a          f      (      x      )        =      a          g      (      x      )          ⟹    f  (  x  )  =  g  (  x  )holds because the exponential function is injective. Therefore, you always need to rewrite both sides of the equation as powers of a before you can equate the exponents.In your situation, you can apply the property:            a      m              a      n        =      a          m      −      n      and thus write      3          5      x      +      2      −      2      (      1      −      x      )        =      3          3      (      4      +      3      x      )      −      6        .Then you can solve the equation  5  x  +  2  −  2  (  1  −  x  )  =  3  (  4  +  3  x  )  −  6.Now, if you&#039;re still wondering why you can&#039;t just equate the exponents of the numerators and those of the denominators, let me show you a simple counterexample to explain why that might not work in general:            3      7              3      5        =            3      8              3      6      holds because if you compute the quotients they are both equal to   9, but   7  ≠  8 and   5  ≠  6.

Seamus Mcknight · Accepted Answer

First subtract the exponents both side and then equate.      3          5      x      +      2      −      2      (      1      −      x      )        =      3          3      (      4      +      3      x      )      −      6        5  x  +  2  −  2  (  1  −  x  )  =  3  (  4  +  3  x  )  −  6    ⟹    x  =  −  3

How do I solve this fractional indices equation (3^(5x+2))/(9^(1-x))=(27^(4+3x))/(729)

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