exponential equation questions

Question

1.Evaluate log₂64+log₃27×log₄1/254

2.solve for x 2log4(x+4)-log₄(x+12)=1

3.solve the equation (cube root256^2)×16^x=64^x-3

4.3x=9^(x^2-1/2) solve for x

Answer 1

1. I think you may have a typo, 256 is a power of 2, but 254 is not.

 

log₂ 64 + (log₃ 27) log₄(1/256)

 

= log₂ 2⁶ + (log₃ 3³) log₄ 4^-4

 

= 6 + (3)(-4)

 

= 6 - 12

 

= -6

 

 

2.

 

2 log₄ (x+4) - log₄ (x+12) = 1

 

log₄ (x+4)² - log₄ (x+12) = 1

 

log₄ [(x+4)²/(x+12)] = 1

 

(x+4)²/(x+12) = 4

 

(x+4)² = 4(x+12)

 

x² + 8x + 16 = 4x + 48

 

x² + 4x - 32 = 0

 

(x - 4)(x + 8) = 0

 

x = 4 or x = -8

 

We can throw out x = -8 since one logarithm in the equation would be log₄ -4, which is undefined. The other choice x = 4, is okay.

 

2 log₄ (x+4) - log₄ (x+12)

 

= 2 log₄ (4+4) - log₄ (4+12)

 

= 2 log₄ 8 - log₄ 16

 

= log₄ 64 - log₄ 16

 

= log₄ (64/16)

 

= log₄ 4

 

= 1

 

 

3. You probably meant to put parentheses around x-3.

 

³√(256²) 16^x = 64^x-3

 

256² (16^x)³ = (64^x-3)³

 

(4⁴)² ((4²)^x)³= ((4³)^x-3)³

 

4⁸ 4^6x = 4^9(x-3)

 

4^6x+8 = 4^9(x-3)

 

9(x - 3) = 6x + 8

 

9x - 27 = 6x + 8

 

3x = 35

 

x = 35/3

 

Check:

 

LHS: ³√(256²) 16^35/3 = 256^2/3 16^35/3 = 4^8/3 4^70/3 = 4^78/3 = 4²⁶

 

RHS: 64^{35/3 - 3} = 64^(35-9)/3= 64^26/3 = 4^78/3 = 4²⁶

 

They match so x = 35/3 works.

 

 

4. As written it is

 

3x = 9^{(x^2 - 1/2)}

 

Usually solutions can't be given in closed form when some occurrences of x are in exponents and some are not. However, by inspection, x = 1 is a solution.

 

The other solution, x = 0.114349699058706... has no closed form.