Solve 5(2x) = 7(x−1). Please help with this algebra II problem. It s solving exponential equations with unequal bases
4 Answers
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When typing algebraic expressions, use ^ before an exponent. It sounds like you mean:
5^(2x) = 7^(x-1)
To solve, take logarithms on both sides. The base doesn’t matter…they just have to be the same base. Use what’s easiest on your calculator.
log[5^(2x)] = log[7^(x-1)]
2x log 5 = (x-1) log 7 = x log 7 – log 7
(2 log 5 – log 7)x = – log 7
x = – (log 7) / (2 log 5)
Simplify that however you like, or get a decimal approximation on a calcuator.
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5(2x) = 7(x – 1)
10x = 7x – 7
3x = – 7
x = – 7/3.
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5(2x) = 7(x − 1)
10x – 7x = -7
3x = -7
x = -7/3
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First distribute:
10x = 7x – 7
Then combine like terms. Terms with x^1 or just simply x go together and terms without a variable go together. So 10x and 7x need to be combined. Subtract 7x from the right side.
10x – 7x= -7
3x=-7
Now isolate the variable x. Divide both sides by 3.
3x/3 =-7/3
x = -7/3
Boom! Done! 😀
Source(s): College student who’s taken algebra II in the past.
