# Log Help A Level Maths repreturns

Watch
Announcements
#1
Question is this: 4^5+x = 20^5
log 4^5+x = log20 ^5
no clue where to go from here
0
2 years ago
#2
(Original post by Pakora99)
Question is this: 4^5+x = 20^5
log 4^5+x = log20 ^5
no clue where to go from here
Check your rules for manipulating logs.

What's equal to?

And use brackets as required, to remove any ambiuity in your posting. log 4^5+x could be log(4^(5+x)), or log(4^5) + x
0
#3
(Original post by ghostwalker)
Check your rules for manipulating logs.

What's equal to?

And use brackets as required, to remove any ambiuity in your posting. log 4^5+x could be log(4^(5+x)), or log(4^5) + x
log a^b = log a + log b?
Its log (4^(5+x))
0
2 years ago
#4
(Original post by Pakora99)
log a^b = log a + log b?
'Fraid not. You do need to know the 3 rules for manipulating logs.

In this case: 0
#5
(Original post by ghostwalker)
'Fraid not. You do need to know the 3 rules for manipulating logs.

In this case: 5+x(log 4) = 5 log 20
5 log 4 + x log 4 = 5 log 20
x log 4 = 5 log 20 - 5 log 4
x= (5 log 20 - 5 log 4) / log 4
Is this it
I just remembered this rule
0
2 years ago
#6
(Original post by Pakora99)
5+x(log 4) = 5 log 20
5 log 4 + x log 4 = 5 log 20
x log 4 = 5 log 20 - 5 log 4
x= (5 log 20 - 5 log 4) / log 4
Is this it
I just remembered this rule
Yep, and you just need your calculator to work that out.
1
#7
Thanks man for the help
(Original post by ghostwalker)
Yep, and you just need your calculator to work that out.
0
2 years ago
#8
(Original post by Pakora99)
5+x(log 4) = 5 log 20
5 log 4 + x log 4 = 5 log 20
x log 4 = 5 log 20 - 5 log 4
x= (5 log 20 - 5 log 4) / log 4
Is this it
I just remembered this rule
5 log 20 - 5 log 4 can be further simplified: remember that log a - log b = log (a/b).
0
2 years ago
#9
(Original post by Pakora99)
5+x(log 4) = 5 log 20
5 log 4 + x log 4 = 5 log 20
x log 4 = 5 log 20 - 5 log 4
x= (5 log 20 - 5 log 4) / log 4
Is this it
I just remembered this rule
Persoanlly (I'm not certain if this is right so let me know haha) but I would've done:

4^(5+x) = 20^5
Log(4^x+5) = log20^5
NOW USING THE RULE THE PERSON ABOUT MENTIONED, you bring the power down in front of the log so:

(X+5)(log 4)= 5Log20
Then divide both by log 4 (can't take away as you did as multiplication)
X+5 = 5log20/log4
Then take away the 5 from both side and you get what x is 0
2 years ago
#10
(Original post by OlaPaw)
Persoanlly (I'm not certain if this is right so let me know haha) but I would've done:

4^(5+x) = 20^5
Log(4^x+5) = log20^5
NOW USING THE RULE THE PERSON ABOUT MENTIONED, you bring the power down in front of the log so:

(X+5)(log 4)= 5Log20
Then divide both by log 4 (can't take away as you did as multiplication)
X+5 = 5log20/log4
Then take away the 5 from both side and you get what x is "Taking logs of both sides" is pretty much always an inefficient approach. The only reason it is used is that back in the day, most calculators couldn't do logs to an arbitrary base (they could only do log base 10 and log base e, or ln), so you would take logs base 10 of both sides. Anyway, simply apply the definition of log to get 5 + x = log base 4 of (20^5), so x = log base 4 of (20^5) - 5.
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Should there be a new university admissions system that ditches predicted grades?

No, I think predicted grades should still be used to make offers (637)
33.44%
Yes, I like the idea of applying to uni after I received my grades (PQA) (805)
42.26%
Yes, I like the idea of receiving offers only after I receive my grades (PQO) (377)
19.79%
I think there is a better option than the ones suggested (let us know in the thread!) (86)
4.51%