# What is the solution of the equation 7^(3x-2)=17 in terms of logarithms?

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We know that log ab = log a + log b and log a^b = b*log a.

Here these are used on 7^(3x-2) = 17.

7^(3x-2) = 17

Take the log of both the sides

=> log [ 7^(3x-2)] = log 17

=> (3x - 2)*log 7 = log 17

=> 3x - 2 = log 17 / log 7

=> 3x = (log 17 / log 7) + 2

=> x = [(log 17 / log 7) + 2]/3

**Therefore x = [(log 17 / log 7) + 2]/3**

7^(3x-2) = 17

First we will apply the logarithm for both sides.

==> log 7^(3x-2) = log 17.

Now we will use the logarithm properties to solve for x.

We know that log a^b = b*log a.

==> log 7^(3x-2) = (3x-2)*log 7.

Let us substitute into the equation.

==> (3x-2)*log 7 = log 17.

Now we will divide by log 7.

==> (3x-2) = log 17 / log 7.

Now we will add 2 to both sides.

==> 3x = (log 17/log 7) + 2

Now we will divide by 3.

**==> x = [(log 17/log 7) + 2 ] / 3**

We'll take decimal logarithms on both sides:

7^(3x-2)=17

log 7^(3x-2)=log 17

We'll use the power rule of logarithms:

(3x-2)*log 7 = log 17

We'll divide by log7 both sides of the equation:

(3x-2) = log 17 / log 7

We'll add 2 both sides:

3x = (log 17 / log 7) + 2

We'll divide by 3:

x = [(log 17 / log 7) + 2]/3

x = (log 17 + 2log 7)/3

x = (log17 + log 7^2)/3

We'll apply product rule:

x = (log 17*49)/3

**x = (log 833)/3**

or

**x = 2.9206**