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Find all values of t such that \( 6^{3t-1} =36^{t-3}\)

 Jul 7, 2020
 #1
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Solve for t:
6^(3 t - 1) = 36^(t - 3)

Take the natural logarithm of both sides and use the identity log(a^b) = b log(a):
log(6) (3 t - 1) = log(36) (t - 3)

Expand out terms of the left hand side:
3 log(6) t - log(6) = log(36) (t - 3)

Expand out terms of the right hand side:
3 log(6) t - log(6) = log(36) t - 3 log(36)

Subtract t log(36) - log(6) from both sides:
(3 log(6) - log(36)) t = log(6) - 3 log(36)

Divide both sides by 3 log(6) - log(36):

t = (log(6) - 3 log(36))/(3 log(6) - log(36)) = - 5

 Jul 7, 2020
 #2
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We only need Law of Indices for this one.

 

\(6^{3t - 1} = 36^{t - 3}\\ 6^{3t - 1} = (6^2)^{t - 3}\\ 6^{3t- 1} = 6^{2(t - 3)}\\ 6^{3t - 1} = 6^{2t - 6}\\ \)

Now, we equate the powers:

 

\(3t - 1 =2t - 6\)

 

You can proceed by solving the linear equation.

 Jul 7, 2020

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