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

Question

I'm supposed to solve using logrithims

Answer 1

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 log_auⁿ = n log_au

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

3/x = (2 - x)(log5/log7)              change of base formula log_ax = 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.

 

Answer 2

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

 

Answer 3

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 log_b x^y = y log_b x. Further simplification results in a quadratic equation. Here I solve the quadratic by completing the square.

7^3/x = 5^2-x

log₅ 7^3/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)