# Log Help A Level Maths repreturns

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Question is this: 4^5+x = 20^5

log 4^5+x = log20 ^5

no clue where to go from here

log 4^5+x = log20 ^5

no clue where to go from here

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#2

(Original post by

Question is this: 4^5+x = 20^5

log 4^5+x = log20 ^5

no clue where to go from here

**Pakora99**)Question is this: 4^5+x = 20^5

log 4^5+x = log20 ^5

no clue where to go from here

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

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(Original post by

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

**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

Its log (4^(5+x))

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#4

(Original post by

log a^b = log a + log b?

**Pakora99**)log a^b = log a + log b?

In this case:

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(Original post by

'Fraid not. You do need to know the 3 rules for manipulating logs.

In this case:

**ghostwalker**)'Fraid not. You do need to know the 3 rules for manipulating logs.

In this case:

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

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#6

(Original post by

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

**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

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Thanks man for the help

(Original post by

Yep, and you just need your calculator to work that out.

**ghostwalker**)Yep, and you just need your calculator to work that out.

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#8

**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

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#9

**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

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

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#10

(Original post by

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

**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

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