Asked by Moku
Solve the following exponential equation
5^2-x=7^5x+7
5^2-x=7^5x+7
Answers
Answered by
Bosnian
5 ^ 2 = 25
7 ^ 5 = 16,807
5 ^ 2 - x = 7 ^ 5 x + 7
25 - x = 16,807 x + 7 Add x to both sides
25 - x + x = 16,807 x + 7 + x
25 = 16,808 x + 7 Subtract 7 to both sides
25 - 7 = 16,808 x + 7 - 7
18 = 16,808 x Divide both sides by 16,808
18 / 16,808 = 16,808 x / 16,808
18 / 16,808 = x
2 * 9 / ( 2 * 8,404 ) = x
9 / 8,404 = x
x = 9 / 8,404
7 ^ 5 = 16,807
5 ^ 2 - x = 7 ^ 5 x + 7
25 - x = 16,807 x + 7 Add x to both sides
25 - x + x = 16,807 x + 7 + x
25 = 16,808 x + 7 Subtract 7 to both sides
25 - 7 = 16,808 x + 7 - 7
18 = 16,808 x Divide both sides by 16,808
18 / 16,808 = 16,808 x / 16,808
18 / 16,808 = x
2 * 9 / ( 2 * 8,404 ) = x
9 / 8,404 = x
x = 9 / 8,404
Answered by
Jai
If you mean,
5^(2-x) = 7^(5x + 7)
Then, we take the ln of both sides
(2-x)*ln(5) = (5x + 7)*ln(7)
1.609*(2-x) = 1.946*(5x+7)
Solving for x,
x = -0.9174
Hope this helps :3
5^(2-x) = 7^(5x + 7)
Then, we take the ln of both sides
(2-x)*ln(5) = (5x + 7)*ln(7)
1.609*(2-x) = 1.946*(5x+7)
Solving for x,
x = -0.9174
Hope this helps :3
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