# Solve 5(2x) = 7(x−1). (Algebra II)?

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Solve 5(2x) = 7(x−1). Please help with this algebra II problem. It s solving exponential equations with unequal bases

• 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.

• 5(2x) = 7(x – 1)

10x = 7x – 7

3x = – 7

x = – 7/3.

• 5(2x) = 7(x − 1)

10x – 7x = -7

3x = -7

x = -7/3

• 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.

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