Suppose we have the following problem to solve:

3^(2x + 1) = 27^(3x)

We can solve this by getting a common base and setting the exponents equal to each other. We know that 3^3 = 27, therefore..

3^(2x + 1) = 3^(9x)

Now that the bases are the same (3), we set the exponents equal to each other and solve for x.

2x + 1 = 9x

1 = 7x

1/7 = x

We can also solve this problem by taking the log of both sides and using rules for logarithms.

log(3^(2x + 1)) = log27^3x

(2x + 1)log3 = (3x)log27

(2x + 1)/3x = log27/log3

(2x + 1)/3x = 3

9x = 2x + 1

1 = 7x

1/7 = x

In this problem is was simpler to get a common base. Sometimes it's quite difficult to get a common base. In those cases it's easier to take the log of both sides first.

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