Next step: Since 8 = 2^3 (2^3)^x + 2^x = 150. Since (a^m)^n = a^(m n) = a^nm = (a^n)^m, (2^3)^x = (2^x)^3 Call (2^x) = a Then, a^3+a = 130 eqn [1] I look for such a combination on RHS (Rght Hand Side), that is 130 = a^3+a as a necessary step before "factorising" the terms on LHS. (2^3)^x + 2^x = (2^x)^3 + (2^x)^1 = a^3 + a^1. Sum of indices (of 2^x) as well as a 4. We need to write RHS as that. 130 = 5 . 26 = 5. (25+1) = (5.25)+5 = 5^3 + 5^1 (the indices add up to 4). Substituting this on RHS a^3+a = 130 = 5^3 + 5^1 a^3+a = 5^3 + 5^1. Bringing on LHS (by adding -(5^3 + 5^1) on both sides), a^3+a -(5^3 + 5^1) = 0 (a^3 - 5^3) + (a^1 - 5^1) = 0 (a^3 - 5^3) + (a - 5) = 0 Since (x^3 - y^3)= (x -y)(x^2 - xy + y^2)in general; write it in the expression (a -5)(a^2 +5a + 5^2) + (a -5) . Take (a -5) as common (a -5)(a^2+5a + 5^2 + 1) = 0 (a -5)(a^2 +5a + 26) = 0. Since Solutions: (i) a-5=0 ⇒ a=5. Recall a = 5 = 2^x Hence, taking log, on both sides, log5 = x log2 x = log5/log2 = log'2(5) = 2.3219 ..read as "log 5 to base 2". (ii) (a^2 +5a + 26) = 0 which is a quadratic equation & roots are not real & a1, a2 = ½[+5±√((5^2) - 4.1.26)] =½[5 ±√(25 - 104)] =½[5 ±√-99] =½[5 ±√(-1)(9)(11)] = ½[5 ±√(-1)√(9)√(11)] a1, a2 = ½[5+ i.3.√11], ½[5 - i.3.√11] Recall a =2^x Taking log on both sides x log2 = ½[5+ i.3.√11], ½[5 - i.3.√11] x1 = log(2^-1)[5+ i.3.√11] x2 = log(2^-1)[5 - i.3.√11] x1 = (-1)log2+ log[5+ i.3.√11] divide by log2 x1 = -1+ log'2[5+ i.3.√11] Similarly x2 = -1+ log'2[5 - i.3.√11]
Next step:
Since 8 = 2^3
(2^3)^x + 2^x = 150. Since (a^m)^n = a^(m n) = a^nm = (a^n)^m,
(2^3)^x = (2^x)^3
Call (2^x) = a
Then, a^3+a = 130 eqn [1]
I look for such a combination on RHS (Rght Hand Side), that is
130 = a^3+a as a necessary step before "factorising" the terms on LHS.
(2^3)^x + 2^x = (2^x)^3 + (2^x)^1 = a^3 + a^1.
Sum of indices (of 2^x) as well as a 4. We need to write RHS as that.
130 = 5 . 26 = 5. (25+1) = (5.25)+5 = 5^3 + 5^1 (the indices add up to 4). Substituting this on RHS
a^3+a = 130 = 5^3 + 5^1
a^3+a = 5^3 + 5^1. Bringing on LHS (by adding -(5^3 + 5^1) on both sides),
a^3+a -(5^3 + 5^1) = 0
(a^3 - 5^3) + (a^1 - 5^1) = 0
(a^3 - 5^3) + (a - 5) = 0
Since (x^3 - y^3)= (x -y)(x^2 - xy + y^2)in general; write it in the expression
(a -5)(a^2 +5a + 5^2) + (a -5) . Take (a -5) as common
(a -5)(a^2+5a + 5^2 + 1) = 0
(a -5)(a^2 +5a + 26) = 0. Since
Solutions: (i) a-5=0 ⇒ a=5.
Recall a = 5 = 2^x
Hence, taking log, on both sides,
log5 = x log2
x = log5/log2 = log'2(5) = 2.3219 ..read as "log 5 to base 2".
(ii) (a^2 +5a + 26) = 0 which is a quadratic equation
& roots are not real & a1, a2 = ½[+5±√((5^2) - 4.1.26)] =½[5 ±√(25 - 104)] =½[5 ±√-99] =½[5 ±√(-1)(9)(11)] = ½[5 ±√(-1)√(9)√(11)]
a1, a2 = ½[5+ i.3.√11], ½[5 - i.3.√11]
Recall a =2^x
Taking log on both sides
x log2 = ½[5+ i.3.√11], ½[5 - i.3.√11]
x1 = log(2^-1)[5+ i.3.√11]
x2 = log(2^-1)[5 - i.3.√11]
x1 = (-1)log2+ log[5+ i.3.√11] divide by log2
x1 = -1+ log'2[5+ i.3.√11]
Similarly x2 = -1+ log'2[5 - i.3.√11]
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