Problem 1 :
Solve for x :
4^{x - 5} = 16
Problem 2 :
Solve for x :
3^{2x - 4} = 9^{3x + 6}
Problem 3 :
Solve for m :
16^{3m - 6} = 1
Problem 4 :
Solve for x. Round your answer to the nearest ten-thousandth.
3^{x} = 17
Problem 5 :
Solve for y. Round your answer to the nearest ten-thousandth.
20^{y} = 56
Problem 6 :
Solve for k. Round your answer to the nearest ten-thousandth.
5 ⋅ 18^{6k} = 26
Problem 7 :
Solve for x. Round your answer to the nearest ten-thousandth.
e^{x - 1} - 5 = 5
Problem 8 :
Solve for p. Round your answer to the nearest ten-thousandth.
-e^{6 - 9p} + 5 = -48.4
Problem 9 :
Solve for r.
8^{r + 4} ⋅ 4^{2r - 10} = 16^{r + 1}
Problem 10 :
A^{x + B} = C^{Dx}
Given that A, B, C and D are all real numbers greater than zero. Solve for x in terms of A, B, C and D.
1. Answer :
4^{x - 5} = 16
The base on the left side is 4 and that of on the right side is 16.
2 = 2^{2}
16 = 2^{4}
Since 2 and 16 are the powers of 2, we can take logarithm with base 2 on both sides of the given equations and solve for x.
log_{2}(4^{x - 5}) = log_{2}(16)
log_{2}[(2^{2})^{x - 5}] = log_{2}(2^{4})
log_{2}(2^{2(}^{x - 5)}) = 4log_{2}(2)
2(x - 5)log_{2}(2) = 4log_{2}(2)
2(x - 5)(1) = 4(1)
2x - 10 = 4
2x = 14
x = 7
2. Answer :
3^{2x - 4} = 9^{3x + 6}
The base on the left side is 3 and that of on the right side is 9.
3 = 3^{1}
9 = 3^{2}
Since 3 and 9 are the powers of 3, we can take logarithm with base 3 on both sides of the given equations and solve for x.
log_{3}(3^{2x - 4}) = log_{3}(9^{3x + 6})
(2x - 4)log_{3}(3) = (3x + 6)log_{3}(9)
(2x - 4)(1) = (3x + 6)log_{3}(3^{2})
(2x - 4)(1) = (3x + 6) ⋅ 2log_{3}(3)
2x - 4 = (3x + 6) ⋅ 2(1)
2x - 4 = 6x + 12
-4x = 16
x = -4
3. Answer :
16^{3m - 6} = 1
The base on the left side is 16 and that of on the right side is 1.
16 = 4^{2}
1 = 4^{0}
Since 16 and 1 are the powers of 4, we can take logarithm with base 4 on both sides of the given equations and solve for m.
log_{4}(16^{3m - 6}) = log_{1}(1)
(3m - 6)log_{4}(16) = 0
(3m - 6)log_{4}(4^{2}) = 0
(3m - 6) ⋅ 2log_{4}(4) = 0
(3m - 6) ⋅ 2(1) = 0
6m - 12 = 0
6m = 12
m = 2
4. Answer :
3^{x} = 17
The base on the left side is 3 and that of on the right side is 17.
Since 3 and 17 are the not the powers of a common number, we can take common logarithm (with base 10) on both sides of the given equations and solve for x.
ln(3^{x}) = ln(17)
x ⋅ ln(3) = ln(17)
Using calculator,
x ≈ 2.5789
5. Answer :
20^{y} = 56
The base on the left side is 20 and that of on the right side is 56.
Since 20 and 56 are the not the powers of a common number, we can take natural logarithm on both sides of the given equations and solve for y.
ln(20^{y}) = ln(56)
y ⋅ ln(20) = ln(56)
Using calculator,
y ≈ 1.3437
6. Answer :
5 ⋅ 18^{6k} = 26
Divide both sides by 5.
18^{6k} = 5.2
The base on the left side is 18 and that of on the right side is 5.2.
Since 18 and 5.2 are the not the powers of a common number, we can take natural logarithm both sides of the given equations and solve for k.
ln(18^{6k}) = ln(5.2)
(6k)ln(18) = ln(5.2)
k ≈ 1.3437
7. Answer :
e^{x - 1} - 5 = 5
Add 5 to both sides.
e^{x - 1} = 10
The base on the left side is e and that of on the right side is 10.
Since e and 10 are the not the powers of a common number, we can take natural logarithm on both sides of the given equations and solve for x.
ln(e^{x - 1}) = ln(10)
(x - 1)ln(e) = ln(10)
(x - 1)(1) = ln(10)
x - 1 = ln(10)
x = ln(10) + 1
x ≈ 3.3026
8. Answer :
-e^{6 - 9p} + 5 = -48.4
Subtract 5 from both sides.
-e^{6 - 9p} = -53.4
Multiply both sides by -1.
e^{6 - 9p} = 53.4
The base on the left side is e and that of on the right side is 53.4.
Since e and 53.4 are the not the powers of a common number, we can take natural logarithm on both sides of the given equations and solve for x.
ln(e^{6 - 9p}) = ln(53.4)
(6 - 9p)ln(e) = ln(53.4)
(6 - 9p)(1) = ln(53.4)
6 - 9p = ln(53.4)
-9p = ln(53.4) - 6
p ≈ 0.2247
9. Answer :
8^{r + 4} ⋅ 4^{2r - 10} = 16^{r + 1}
8 = 2^{3}
4 = 2^{2}
16 = 2^{4}
Since the bases 8, 4 and 16 are the powers of 2, we can take logarithm with base 2 on both sides of the given equations and solve for r.
log_{2}(8^{r + 4} ⋅ 4^{2r - 10}) = log_{2}(16^{r + 1})
log_{2}(8^{r + 4}) + log_{2}(4^{2r - 10}) = log_{2}(16^{r + 1})
(r + 4)log_{2}(8) + (2r - 10)log_{2}(4) = (r + 1)log_{2}(16)
(r + 4)log_{2}(2^{3}) + (2r - 10)log_{2}(2^{2}) = (r + 1)log_{2}(2^{4})
(r + 4) ⋅ 3log_{2}(2) + (2r - 10) ⋅ 2log_{2}(2) = (r + 1) ⋅ 4log_{2}(2)
(r + 4) ⋅ 3(1) + (2r - 10) ⋅ 2(1) = (r + 1) ⋅ 4(1)
3(r + 4) + 2(2r - 10) = 4(r + 1)
3r + 12 + 4r - 20 = 4r + 4
7r - 8 = 4r + 4
3r = 12
r = 4
10. Answer :
A^{x + B} = C^{Dx}
Since the bases A and C are alphabets, we can take natural logarithm on both sides of the given equations and solve for x in terms A, B, C and D.
ln(A^{x + B}) = ln(C^{Dx})
(x + B)ln(A) = (Dx)ln(C)
xln(A) + Bln(A) = Dxln(C)
xln(A) - Dxln(C) = -Bln(A)
x[ln(A) - Dln(C)] = -Bln(A)
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