How do we solve 6^3t-1=5 for #t#?

Answer 1

To solve the equation 6^(3t - 1) = 5 for t, you first need to isolate the exponent term by taking the logarithm of both sides of the equation with base 6. This yields:

3t - 1 = log₆(5)

Then, you solve for t by isolating it:

3t = log₆(5) + 1

Finally, divide both sides by 3:

t = (log₆(5) + 1) / 3

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Answer 2

Ethan Adkins

#t=1/3log_6(30)#

Without the appropriate formatting, I think it's

#6^(3t-1)=5#. If so

#log_(6)5=3t-1# or

#3t=log_(6)5+1=log_(6)5+log_(6)6=log_(6)(5xx6)#

#3t=log_6(30)#

Hence, #t=1/3log_6(30)#

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Answer 3

Avery Alexander

For #6^(3t)-1=5#, the answer is #t = 1/3#.
For #6^(3t-1)=5#, the answer is #t=(1+log 5/log 6)/3#.

If it is #6^(3t)-1=5# then #6^(3t)=6#.

So, #3t = 1 and t = 1/3#.

For #6^(3t-1)=5#, equate logarithms.

#(3t-1) log 6 = log 5#. Solving,

#t=(1+log 5/log 6)/3#.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.