alcazar

iTrader: (2)

Logarithms this time. I don't remember doing anything with logs except how to manipulate them at grammar school.

So:

1. e^4-x = 3e^x. Evaluate.

2. 3e^-x+2 = 5e^x-1. Evaluate.

3. 2^x - 2^-x = 6. Evaluate.


Explanations, or links gratefully received. TIA.

speedking

1. take logs of both sides, 4-x = ln3 + x
add x to both sides and deduct ln3
2x = 4 - ln3
x = 2 - (ln3)/2 = 1.4507

alcazar

iTrader: (2)

Thankyou, that makes sense.
You can do the second one the same way, then, but what of the third?

scud8

iTrader: (1)

Just take log2 of both sides (after a bit of rearrangement). You can do logs with any base, 10, e (a natural log) or 2 (often used to work out the complexity of computer algorithms).

Boro

iTrader: (1)

Yeh, what he said :-p

blackpoolrock

thats how i was going to explain it !!!!! OMG !!!!!

speedking

e.g. 10² = 100
log base 10 of 100 =2

so 2^x = y
log base 2 of y = x

HTH