math question?

Question

This problem is driving me crazy... how would you do...

log2 (3x + 2) - log4 x = 3

Since the bases aren't the same, I don't know what to do >.<

Can someone help me figure out the steps so I can know what I'm doing?... I tried somethin and got 14/3... But it didn't work.

Answer 1

Make the bases the same.
If log_4 x = a, say
then 4^a = x
and 2^(2a) = x
log_2 x = 2a
a = ½* log_2 x = log_4 x
So log_2(3x+2) - ½log_2 x = 3
2 log_2(3x+2) - log_2 x = 6
log_2 (3x+2)² - log_2 x = 6
log_2 ( (3x+2)² / x) = 6
(3x+2)²/x = 64
9x² + 12x + 4 = 64x.
9x² -52x + 4 = 0.
x = 1/18[ 52 + √2560] = 1/18[52 + 16√10] = 1/9[26 + 8√10]
and x = 1/9[ 26 - 8√10].

Answer 2

the answer is 11. 3x is the same as saying 3 times 3 since x=3. so 3 times 3. then do +2 so 3 times 3 is 9 plus 2 =11 ^^ enjoy hehe

Answer 3

Is that log, to the base 2? , or log of 2(3x+2) ?

if there are no bases given it could well mean Ln ( Log Natrual) in which case you could solve as follows

Log ((2(3x+2))/4x)= 3
(6x+4)/4x= e^3 < solve from there

Sorry if thats utter rubbish- Its just that if I don't get back and see your reply to my first Q I wouldn't be able to show how to solve that situation.

Answer 4

change of base:

log2 (3x +2) - log2 x/log2 4 = 3

log 2 (3x +2) - (log2 x)/2 = 3

log 2 (3x +2) - 1/2log2 x = 3

log2 (3x +2) / sqrt(x) = 3

8 = (3x +2) / sqrt(x)
8 sqrt(x) = 3x + 2

64x = 9x^2 + 12x + 4
9x^2 - 52x +4 = 0

x= [52 +/- 16 sqrt(10) ] /18
x = [26 +/- 8 sqrt(10)]/9

Answer 5

rewrite using change of base formula:

[log(3x + 2) / log(2)] - [log(x) / log(4)] = 3

now using properties of logs, rewrite:

log (3x + 2 - 2) - log (x - 4) = 3
log (3x) - log (x - 4) = 3
log (3x/(x-4)) = 3
3x/(x-4) = 10^3
3x = 10^3(x-4)
3x - (10^3)x = -4*10^3
-997x = -4*10^3
x = 4.0120361

(not bad for a pro mexican wrestler, huh?)

Answer 6

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