Question

Solve for t using logarithms with base a.

\[3a^{4t}=10\]

Answer 1

Use this :

\[Loga^b = b.Loga\]

Answer 2

Okay, I'm going to put all my steps... sorry...

Answer 3

\[log_a(3a^{4t}=log_a(10))\]
\[log_a(3)+log_a(a^{4t})=log_a(10)\]
\[4t=log_a(10)-log_a(3)\]
\[4t=log_a(\frac{10}{3})\]
\[t=\frac{1}{4}(log_a(\frac{10}{3}))\]

Answer 4

What is the answer???

Answer 5

The last line... or do I have to multiply it out...?

Answer 6

How can you multiply it??
I think it will go into the power..

Answer 7

Yeah but either way I still got the wrong answer...

Answer 8

I am asking for what is the correct answer??

Answer 9

Oh it's an even question but wolfram has the answer:

\[t=\frac{log(\frac{10}{3})}{4 log (a)}\]

But is there more than one way to do this because another website also gives has the answer:
\[t=\frac{log_a(10)}{4log_a(3a)}\]

Answer 10

Second one I have got...

Answer 11

Is there more than one to do this... or anyway you do it your supposed to come up with the same answer?

Answer 12

You have made a mistake there..
are you able to point out or not???

Answer 13

No am not...

Answer 14

Why you taken log to the base a??
Well, it is not necessary..

You can take it as log to the base 10 also..

Answer 15

By taking log to the base 10.
you will get:

\[Log(a^{4t}) = 4t.Log(a)\]

Use this you will get your answer..

Answer 16

I show both the methods one by one just wait:

Answer 17

Well, one... those are the instructions and two...Because I thought that you have to use the cancellation law if you have an exponent a^x to you raise it to the log_a to find x or make it equal to whatever, x, is in the exponent.

Answer 18

Okay :).

Answer 19

But the instructions state using logarithms with base a

Answer 20

First Method:

Take Log both the sides:

\[Log(3^{4t}) = Log(10)\]
\[Log(3) + Log(a^{4t}) = Log(10)\]
\[4t.Log(a) = Log(10) - Log(3)\]
\[4t.Log(a) = Log(\frac{10}{3})\]

\[t = \frac{Log(\frac{10}{3})}{4.Log(a)}\]

Answer 21

Yes, second method is using that instruction..Just wait I will explain..

Answer 22

Have you left out the first line of the by accident \[log_a(3a^{4t})\] or...

Answer 23

Take Log(to the base a both sides),

\[Log_a(3^{4t}) = Log_a(10)\]

\[4t.Log_a(3a) = Log_a(10)\]

So, from here evaluate t:

\[t = \frac{Log_a(10)}{4.Log_a(3a)}\]

Here is your second answer..
You can do more here if you want to..
Just carry the 4 to the power of (3a), you will get 81a^4..

Getting or not??

Answer 24

No... for the first one... can't I use the cancellation law that state \[log_a(a^x)=x\] to get rid of the a...

Answer 25

Sorry, in first line it will come 3a^(4t) and not 3^(4t)..

You can go with your method also there is nothing wrong in that...

Answer 26

But... I thought regardless of the method am suppose to get the same answer?

Answer 27

See, the answer provided by wolfram and the other website is also different..
Your answer can also be different..
What is the problem in that???

Answer 28

Sigh, but their answers are very similar...

Answer 29

See, what more can you do is:

\[4.Log_a(3a) = 4.Log_a(3) + 4.Log_a(a) = 4(Log_a(3) + 1)\]

Answer 30

:\

Answer 31

nothing wrong with what you have in your first post....
\(\large t=\frac{1}{4}(log_a(\frac{10}{3})) \)

Answer 32

but you can also utilize the properties of logs:
\(\large t=\frac{1}{4}(log_a(\frac{10}{3})) \)
\(\large t=\frac{1}{4}(log_a10-log_a3) \)
\(\large t=\frac{1}{4}log_a10-\frac{1}{4}log_a3 \)

Answer 33

Or it can be:

\[t = Log_a(10)^{1/4} - Log_a(3)^{1/4}\]

Answer 34

yes^^^ :)

Answer 35

How would I convert that back to the original?

Answer 36

just work backwards from the formula/properties

Answer 37

Okay, I'll put it in another question. Thanks so much for all your help I really do appreciate it everyone!!

Answer 38

but also, just to be clear with what waterineyes wrote, the expression is:
\[\huge t=Log_a(10^\frac{1}{4})-Log_a(3^\frac{1}{4}) \]

Answer 39

@dpaInc how you write this in so large form?
can you please tell???

Answer 40

instead of "\large", use "\huge" in your latex...:)