Solve the equation lg(3^y - 2^4-y) = 2 + 1/3lg8 - 1/4ylg16?

Question

Clear explanations would be appreciated as I just started out on this type of Mathematics and I'm having a bad time with it.

lg = log base 10

Answer 1

log[3^y - 2^(4-y)] = 2 + (1/3)log8 - (1/4)ylog16
log[3^y - 2^(4-y)] = 2 + log[8^(1/3)] - ylog[16^(1/4)]
log[3^y - 2^(4-y)] = 2 + (log 2) - y(log 2)
log[3^y - 2^(4-y)] = 2 + (1 - y)(log 2)
log[3^y - 2^(4-y)] = 2 + log[2^(1 - y)]
10^{log[3^y - 2^(4-y)]} = 10^{2 + log[2^(1 - y)]}
3^y - 2^(4-y) = (10^2){2^(1 - y)}
3^y - 16/(2^y) = (100){2/(2^y)}
(3^y)(2^y) - 16 = 200
6^y = 216
y = 3

Answer 2

Ok, I'll try and make sense :) I'll put the rules I used for each step in square brackets [ ] to the side so that you can see what I'm doing in each step...

lg(3^y - 2^4-y) = 2 + 1/3lg8 - 1/4ylg16
lg(3^y - 2^4-y) = 2 + lg8^1/3 - lg16^1/4y [p logn = logn^p]
lg(3^y - 2^4-y) = 2 + lg((8^1/3)/(16^1/4y)) [logx-logy = log x/y]

Now we know that 8=2^3 and 16=2^4
So that this doesn't get all big and confusing in the middle of your logs equation I'll do the working for simplifying (8^1/3)/(16^1/4y) here before putting it back into the main equation.

so, (2^3)^1/3 / (2^4)^1/4y [(x^a)^b = x^ab]
= 2/2^y
=2^1-y

So now, all that ugly mess of powers and numbers is gone we are left with....

lg(3^y - 2^4-y) = 2 + lg2^1-y
lg(3^y - 2^4-y) = lg100 +lg2^1-y [2=lg100]

therefore, if the logs are the same on both sides then the numbers after the logs must also be the same...

3^y - 2^4-y = 100 + 2^1-y
then I split 2^4-y up into 2^4 x 2^-y. 2^-y is also the same as 1/2^y so you get 2^4 / 2^y
3^y - (16/2^y) = 100+2^1-y [16 = 2^4]
3^y - (16/2^y) = 100+(2/2^y) [same as above only with 2^1-y]

now multiply by 2^y on both sides....
(3^y)(2^y) - 16 = 100(2^y) +2
(3^y)(2^y) = 100(2^y) +18 [+16 on both sides]
(3^y) = 118 [ / 2^y on both sides and then add on 18 to the 100]
lg(base3)118=y

Now you can't do logs that are base anything other than 10 on a calculator (well on mine anyways...) so you need to use the change of base rule: lg(basex)y=lgy/lgx
lg118/lg3 = y [*note that when I don't put in a base it is assumed base10]
Now plug those numbers into a calculator and you get... 4.342 = y
(rounded to 3 decimal places)

I think my maths is right... This time :) I hope I was abe to help! Good luck!

Answer 3

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