# Solve. Round answer to 2 decimal places. 7^(3x) = 8^(2x)

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We have to solve 7^(3x) = 8^(2x)

Now the base of the two terms are not the same, so we use logarithms.

Take the logarithm of both the sides.

log( 7^(3x)) = log(8^(2x))

we use the relation log (a^b) = b*log a

=> 3x * log 7 = 2x * log 8

=> 3x * log 7 - 2x * log 8 = 0

=> x* (3 * log 7 - 2 * log 8) = 0

=> x = 0/ (3 * log 7 - 2 * log 8)

=> x = 0

Therefore the only possible value for x is **x = 0**

We'll take natural logarithms both sides.

ln 7^(3x) = ln 8^(2x)

We'll apply power rule:

3x*ln 7 = 2x*ln 8

We'll subtract 2x*ln 8 both sides:

3x*ln 7 - 2x*ln 8 = 0

We'll factorize by x:

x(3ln7 - 2ln8) = 0

x(ln7^3 - ln8^2) = 0

We'll apply quotient rule:

x[ln(7^3/8^2)] = 0

Since ln(7^3/8^2) is not zero, only the factor x could be zero, so that the product to be zero.

**The solution of the equation is x = 0.**