How do I get the unknown in calculations like these?

Hugin · #1 by **Hugin** » Feb 02, 2011 12:46 am

How do I get the unknown (t) in the example below? Yeah, I know what the answer is, but in that case of problems, which method do you use to get it?

3^t=27

Or this?

1.075^t=1.5

Could someone please help out?

Tinker Grey · #2 by **Tinker Grey** » Feb 02, 2011 1:12 am

log(3t) = log(27)
t * log(3) = log(27)
t = log(27)/log(3)
t = 3

Works with ln, too.

Hugin · #3 by **Hugin** » Feb 02, 2011 12:04 pm

Thanks a lot!

Tinker Grey · #4 by **Tinker Grey** » Feb 02, 2011 2:58 pm

My pleasure.

gmk2 · #5 by **gmk2** » Feb 02, 2011 4:02 pm

Alternatively, most calculators are capable of doing it directly:
3^t=27
t=log327

1.075^t=1.5
t=log1.0751.5

sweitzen · #6 by **sweitzen** » Feb 02, 2011 4:58 pm

Try Wolfram Alpha.

Just input "solve 3^t = 27"

THWOTH · #7 by **THWOTH** » Feb 03, 2011 4:35 pm

sweitzen wrote:Try Wolfram Alpha.

Just input "solve 3^t = 27"

Google can do that for you too : http://goo.gl/5Mo4n

¬ :evilgrin:

Ihavenofingerprints · #8 by **Ihavenofingerprints** » Feb 03, 2011 4:48 pm

I did foundation maths in high school, this is how we roll:

3^t=27
^t=27/3
t= √(27/3)
t=√9
t=3

I think that was it anyway.

Tinker Grey · #9 by **Tinker Grey** » Feb 04, 2011 4:21 am

Ihavenofingerprints wrote:I did foundation maths in high school, this is how we roll:

3^t=27
^t=27/3
t= √(27/3)
t=√9
t=3

I think that was it anyway.

10 ^ t = 100
^t = 100/10
t = √10
t ~= 3.1623

I don't think so ... unless there was some trick (or I'm humor impaired).

Ihavenofingerprints · **#10** by **Ihavenofingerprints** » Feb 04, 2011 7:11 am

Hmm, i've done something wrong. Its been 4 months since i have picked up a maths book so maybe i've re-arranged the (^) wrong.

Darkchilde · **#11** by **Darkchilde** » Feb 04, 2011 5:17 pm

Using logarithms is the fastest and easiest way to do it.

MedGen · **#12** by **MedGen** » Feb 19, 2011 10:40 am

Ihavenofingerprints wrote:Hmm, i've done something wrong. Its been 4 months since i have picked up a maths book so maybe i've re-arranged the (^) wrong.

The '^' means an exponent or power. So x^2 simply means 'x squared' (x2). So 3^t is '3 to the power of t, simply dividing by 't' and rearranging in the way you've set out doesn't solve it.

Calilasseia · **#13** by **Calilasseia** » Jun 13, 2011 12:39 am

By the way, we have superscript and subscript tags on the board for this.

So, we have, for example:

3t = 27

log10[3t] = log1027

t log103 = log1027

t = log1027 ÷ log103

However, log1027 ÷ log103 = log327, by a well-known rule of logarithms, and log327 = 3.

Therefore, t = 3.

Calilasseia · **#14** by **Calilasseia** » Jun 13, 2011 12:42 am

Double post.

Incidentally, you can use any base of logarithms in the above, not just base 10. So you could have used logarithms to the base 3 right from the start. Viz:

3t = 27

log3[3t] = log327 = 3

t log33 = 3

But since log33 = 1, we have t = 3.