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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.
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