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1 February, 19:28

Demonstrate two different ways to solve the equation 5^ (2x+1) = 25.

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  1. 1 February, 21:38
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    Step-by-step explanation:

    Method 1: Taking the log of both sides ...

    So take the log of both sides ...

    5^ (2x + 1) = 25

    log 5^ (2x + 1) = log 25 <- - use property: log (a^x) = x log a ...

    (2x + 1) log 5 = log 25 <- - distribute log 5 inside the brackets ...

    (2x) log 5 + log 5 = log 25 <- - subtract log 5 both sides of the equation ...

    (2x) log 5 + log 5 - log 5 = log 25 - log 5

    (2x) log 5 = log (25/5) <- - use property: log a - log b = log (a/b)

    (2x) log 5 = log 5 <- - divide both sides by log 5

    (2x) log 5 / log 5 = log 5 / log 5 <-- - this equals 1 ...

    2x = 1

    x=1/2

    Method 2

    5^ (2x+1) = 5^2

    2x+1=2

    2x=1

    x=1/2
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