Maths Olympiad | A interesting Exponential Equation Solve for X?

[@ngwakomokgobu9612]

(5^a) + (5^a) + (5^a) =60 Let (5^a) = k k+k+k =60 3k =60 k = 20 Thefore assign the value 20 to variable k. (5^a) = 20 a.ln(5) = ln(20) dividing both sides by the natural log, ln(5), we get : a = 1.861

[@1_Sumit_Solanki]

Let's find exact value of a (5^a) + (5^a) + (5^a) =60 3•(5^a) = 60 5^a = 60/3 5^a = 20 [If a = 1][5^1 = 5] [If a = 2][5^2 = 25] a is in between 1 and 2 [If a = 1.5][5^1.5 ≈ 11.18] [If a = 1.9][5^1.9 ≈ 21.28] [If a = 1.8][5^1.8 ≈ 18.12] a is in between 1.8 and 1.9 [If a = 1.85][5^1.85 ≈ 19.64] [If a = 1.86][5^1.86 ≈ 19.96] [If a = 1.87][5^1.87 ≈ 20.28] a is in between 1.86 and 1.87 [If a = 1.861][5^1.861 ≈ 19.99] [If a = 1.862][5^1.862 ≈ 20.02] a is in between 1.861 and 1.862 [If a = 1.8615][5^1.8615 ≈ 20.00] So, 5^a ≈ 5^ 1.8615 ≈ 20.00 That's why {a = 1.8615 }

[@ArnavGarg-rl6qh]

Just say log5(20)

[@JasonHughes-in7xo]

Did he literally write “1+1+1” ?

[@pas6295]

Correct.

[@pas6295]

Let us first write the equation it is 5^a+5^a+5^a=60 So 5^a=60/3=20. 20 can be written as 5^5-5. 5 ^a=5^2-5. 5^a+5=5^2

[@sriramnivas5728]

a=1.8614 approx.

[@sriramnivas5728]

a = 1.8614 approx.