Which of the following is the solution to the equation 25^(z + 4) = 125?

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Mozelle D&#039;Amore · Accepted Answer

Which of the following is the solution to the equation 25^(z + 4) = 125?Let's solve this by common bases...but to do that, we want to rewrite both sides of the equation with the base being 5. Observe that on the left side 25 can be written 5^2...so that makes the left side...(5^2)^(z + 4) and by the exponent property: (x^a)^b = x^(a*b)..this becomes 5^(2(z + 4))...On the right side, notice that 125 can be written as 5^3...so...25^(z + 4) = 125 5^(2(z + 4)) = 5^3 Now since the bases are the same, the exponents must be the same as well, so set the exponents equal to each other and solve for x. 2 (z + 4) = 3 Apply the distributive property on the left side: a (b + c), multiply everything inside the brackets by the expression in front..2(z) + 2(4) = 3 2z + 8 = 3 Now solve this linear equation by inverse operations (and remember that what you do to one side you do to the other..). Begin by isolating 2z by subtracting 8 both sides of the equation...the 8 and -8 on the left cancels out... 2z + 8 = 3 2z + 8 - 8 = 3 - 8 2z = -5 To isolate z now, simply divide both sides of the equation by 2...the division cancels out the multiplication on the left side.. 2z/2 = -5/2 z = -5/2

Mozelle D&#039;Amore · Answer

If z is a real number:z + 4 = log25z + 4 = ln(125) / ln(25)If you want ALL solutions:z + 4 = (ln(125) + 2ipi*n) / ln(25), for any integer nz = ((ln(125) + 2ipi*n) / ln(25)) - 4, for any integer nz = (ipin/ln(5)) - 2.5, for any integer nz =~ 1.9519812658311713982736523822486in - 2.5, for any integer n

Mozelle D&#039;Amore · Answer

25^(z+4)=125 take natural log of both sides(z+4)ln25=ln125 divide both sides by ln25z+4=ln125/ln25 subtract 4 from both sidesz=(ln125/ln25)-4z=-2.5check..25^(1.5)=125

Mozelle D&#039;Amore · Answer

25 = 5^25^(2(z+4)) = 5^3 2(z+4) =32z +8 = 32z = -5z = -5/2

Carol Funk

Which of the following is the solution to the equation 25^(z + 4) = 125?

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