\text{If }y=\frac{3^x+2}{3^x-2},\text{ then }x=\log_3\frac{2y+2}{y+k}\text{ f

generals336

generals336

Answered question

2021-11-08

If  y=3x+23x2,  then  x=log32y+2y+k  for some real number k. What is the value of k?  

Answer & Explanation

Corben Pittman

Corben Pittman

Skilled2021-11-09Added 83 answers

Step 1
Use the properties of logarithm to simplify the following. Substitute the value of y and solve for k.
Step 2
Simplify the following equation x=log32y+2y+k by using the properties of logarithms. As the base of the logarithm is 3 thus raise the terms to the power of 3 on both the sides then substitute the value of y in it and simplify.
x=log32y+2y+k
3x=3log32y+2y+k
3x=2(3x+23x2+2)(3x+23x22)
3x=2(3x+2)+2(3x2)(3x+2)+k(3x2)
3x[(3x+2)+k(3x2)]=23x+4+23x4
32x+23x+k[32x2(3x)]=43x
k[32x2(3x)]=43x23x32x
k=2(3x)32x32x2(3x)
k=32x2(3x)32x2(3x)
k=1
Thus the value of k is -1.

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