# Equation.Find the solution of the equation 7^(3x-2)=17 in terms of logarithms.

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### 3 Answers

The equation given is 7^(3x - 2) = 17

take the log of both the sides

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

use the property log a^b = b*log a

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

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

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

=> x = 2/3 + (1/3)*(log 17/log 7)

**The solution of the equation is x = 2/3 + (1/3)*(log 17/log 7)**

The equation 7^(3x-2)=17 has to be solved. This can be done by using logarithms to any base. For convenience use a common base like 10.

Take the logarithm to base 10 of both the sides of 7^(3x-2)=17

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

Now use the property log a^b = b*log a

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

Isolate x to the left hand side

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

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

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

The solution of the equation is 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**