# Logs question

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#1
Solve

8^y = 4^(2x+3)
and
log 2y = log 2x + 4

Thanks
0
15 years ago
#2
solve for what? x? y?
0
15 years ago
#3
Solve

8^y = 4^(2x+3)
and
log 2y = log 2x + 4

Thanks
(2^3)y = (2^2)^(2x+3)
2^3y = 2^2(2x+3)
2^3y = 2^(4x+6)
3y=4x+6
y = (4x+6)/3

log2y = log2x + 4

You can now use a substitution to obtain a logarithmic equation in one variable. Transform it into the form log[something] = [something] which you can then solve.
0
15 years ago
#4
Solve

8^y = 4^(2x+3)
and
log 2y = log 2x + 4

Thanks
err... first equation is simple indicies rules...
8^y=2^3y
4^(2x+3)=2^2(2X+3)...

check your indicies rules... then.. equate them...
8^y=4^(2x+3)
2^3y=2^2(2X+3)

Take Log base 2 both sides... 3y=2(2X+3)
So... 3y=4X+6

SECOND EQUATION THERE IS A PROBLEM... do you mean
log 2y = (log 2x) + 4 OR log 2y = log (2x+4)

For FORMER one.... log 2y = (log 2x) + 4
Take power of 10... assuming you mean log base 10...
10^log 2y =10^((log 2x) + 4)
2y = 2X times 10^4
So.... y = 10000X

For LATTER ONE... log 2y = log (2x+4)
you base 10 on both sides...
so 2Y=2X+4.... so Y=X+2

NOW COMBINE FIRST & SECOND RESULTS

For 3y=4X+6 and y = 10000X...
X = 3/14998 and Y= 15000/7999

For 3y=4X+6 and Y=X+2...
X = 0 and Y = 2

MAKE SURE YOU MAKE CLEAR WHAT YOU MEAN... !!!
0
15 years ago
#5
(Original post by anchemis)
err... first equation is simple indicies rules...
8^y=2^3y
4^(2x+3)=2^2(2X+3)...

check your indicies rules... then.. equate them...
8^y=4^(2x+3)
2^3y=2^2(2X+3)

Take Log base 2 both sides... 3y=2(2X+3)
So... 3y=4X+6

SECOND EQUATION THERE IS A PROBLEM... do you mean
log 2y = (log 2x) + 4 OR log 2y = log (2x+4)

For FORMER one.... log 2y = (log 2x) + 4
Take power of 10... assuming you mean log base 10...
10^log 2y =10^((log 2x) + 4)
2y = 2X times 10^4
So.... y = 10000X

For LATTER ONE... log 2y = log (2x+4)
you base 10 on both sides...
so 2Y=2X+4.... so Y=X+2

NOW COMBINE FIRST & SECOND RESULTS

For 3y=4X+6 and y = 10000X...
X = 3/14998 and Y= 15000/7999

For 3y=4X+6 and Y=X+2...
X = 0 and Y = 2

MAKE SURE YOU MAKE CLEAR WHAT YOU MEAN... !!!
Indeed!

log 2y = (log 2x) + 4 is a fair bit more tricky to deal with than

log 2y = log (2x+4) !!

Argh!
Aitch
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