How do you solve log base 16(2x-3)=log base 4(x+2)?

Expert Answers

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We have to solve log(16) ( 2x - 3) = log(4) ( x + 2) for x.

log(16) ( 2x - 3) = log(4) ( x + 2)

=> log(4)(2x - 3) / log(4) 16 = log(4) ( x + 2)

=> log(4)(2x - 3) / log(4) 4^2 = log(4) ( x + 2)

=> log(4)(2x - 3) / 2* log(4) 4 = log(4) ( x + 2)

=> log(4)(2x - 3) / 2 = log(4) ( x + 2)

=> log(4)(2x - 3) = 2*log(4) ( x + 2)

=> log(4)(2x - 3) = log(4) ( x + 2)^2

Now we can equate 2x - 3 and (x + 2)^2

(x + 2)^2 = 2x - 3

=> x^2 + 4 + 4x = 2x - 3

=> x^2 + 2x + 7 = 0

But we see that 2^2 < 4*7

We get only complex values of x when the equation is solved and the log of a complex number is not defined. There are no valid solutions of x.

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