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# 7^(3/x)=5^(2-x)

I'm supposed to solve using logrithims

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### 3 Answers

Hello Susie,

(7)3/x = 5(2 - x)                         original equation

log7(7)3/x = log7(5)(2 - x)             Take log (base 7) of each side.

(3/x)log77 = (2 - x)log75            Use logaun = n logau

(3/x) * 1 = (2 - x)log75              logaa = 1 because a1 = a

3/x = (2 - x)(log5/log7)              change of base formula logax = log x/log a

3/x = (2 - x)(0.6989/0.8451)       use calculator to find the values of log5 and log7

3/x = (2 - x)(0.827)

3 = (2 - x)(0.827)(x)                  Multiply by x on both sides

3/0.827 = (2 - x)(x)                   divide by 0.827 on both sides

3.628 = 2x - x2                                 distribute x on right side

Finally, it is a quadratic equation

x2 - 2x + 3.628 = 0

Here a = 1, b = -2 and c = 3.628

Use the quadratic formula and solve for x. You'll get two values of x as an answer. Just plug in the values of a, b, and c in the formula.

x = (-b ± √(b2 - 4ac))/2a

Good luck. If you need any help you can ask.

Take ln of each side, (3/x)ln 7 =  (2-x)ln 5

Rearrange, (x^2)/3   -   (2x)/3    +    (ln 7)/(ln 5)  =0

This will solve as a 2nd degree polynomial with a=1/3    b=2/3    c=(ln7)/(ln5).

Due to the discriminant, (b^2 - 4ac) being negative this will have non real roots (i.e., it will involve (√-1) or i.  Notice both roots are complex and non real or imaginary.

Solve the equation for x = 1 ± 1.62i

### Comments

Roman's attack is more elegant, in that it avoids use of the quadratic formula.  Completing the square is easier.  The end result is the same.  Seeing more than one attack of a problem can make a problem clearer.

Take the logarithm of both sides. In this case, base 5 logarithm is the best choice. Then simplify using properties of logarithms, in this case logb xy = y logb x. Further simplification results in a quadratic equation. Here I solve the quadratic by completing the square.

73/x = 52-x

log5 73/x = 2 - x

(3/x) log5 7 = 2 - x

3 log5 7 = 2x - x2

x2 - 2x + 3 log5 7 = 0

x2 - 2x + 1 = 1 - 3 log5 7

(x - 1)2 = 1 - 3 log5 7

x - 1 = ±√(1 - 3 log5 7)

x = 1 ± √(1 - 3 log5 7)

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