Question

Agh can someone help pls

2+log5 of 10 + log5 of 2 - 5

Having a blank moment :(

Answer 1

Well, for starters we can simplify 2-5 (I'm sure you can do that easily) and then we look at \[\log_{5} 10 + \log_{5}2 \] Using a certain log law, you could turn that into one log expression.

Answer 2

so i get -3log5 of 20??

Answer 3

Is your question in the form \[2 + \log_{5}10 + \log_{5}2 - 5? \] If so, the -3 and the log 5 of 20 would be two separate terms:
\[-3 + \log_{5}20 \] or \[\log_{5}20 - 3\]

Answer 4

ah ok and no further simplification?

Answer 5

I don't think so - you could try it though. Using the change of base rule, you could put \[\frac{ \ln 20 }{ \ln5 }\] to evaluate the log, then subtract 3, but your answer will just be a decimal approximation. If I was you, I would try it in all cases anyway, because sometimes it evaluates to an exact integer, which could be useful.

Answer 6

For example, if you had \[\log_{4}16 \] and didn't immediately see what that could evaluate to, you could check it out using the change of base rule and discover it evaluates to 2.

Answer 7

\[\log_{3}x^5+\log_{3}x^2-\log_{3}x^4 \]

Answer 8

is the answer log3x^6??

Answer 9

\[\log_{3}x^6 \]

Answer 10

Consider what you're doing here. When you add the logs to turn them into one log, you're multiplying the inside of the logs together (and when you're subtracting, you divide by the inside of the log). In this case, you have \[\log_{3} \frac{ x^5 \times x^2 }{ x^4 }\]

Answer 11

What should that evaluate to be?

Answer 12

\[\log_{3}x^3 \]

Answer 13

That looks correct to me! You could also use the rule that \[\log_{b} x^n = nlog_{b}x\] to get 3 outside of log 3 of x, as your answer.

Answer 14

cool

Answer 15

\[3^{2x}-12*3^{x}+27=0\]

Answer 16

This type of question involves a substitution. I'll give you the general structure to work these out.
1. Let u = *something that can be represented as a variable of a quadratic*
2. Solve the quadratic in terms of u.
3. Allow the u to become that *something* again, and you'll usually have a couple of equations.
4. Solve these equations for x, if possible.

Answer 17

ok thanks for that

what about
3^x = 30 solve for x

Answer 18

Use the change of base rule:
\[b^x = y\]
\[x = \frac{ \ln y }{ \ln b }\]

Answer 19

A way I like to remember it is "base on the bottom", in case you forget which number goes on the bottom and which goes on the top.

Answer 20

I am still lost on the previous one Would I say let 3^2x = y??? and therefore 3^x also = y

Answer 21

or is it 3^x = y??

Answer 22

By previous one, do you mean the one involving a substitution and quadratic? If so, consider \[ 3^{2x} = \left( 3^x \right)^2\]

Answer 23

so therefore if I say let 3x = y then I would get y^3 - 12 =27?

Answer 24

If you let 3x = y, you would get y^2 -12y + 27 = 0, and you could then solve that as a quadratic.

Answer 25

ahhhh ok thanks so much :)

Answer 26

\[\log_{4} (0.5x+1)=2\] how do I go about this?

Answer 27

4^2 = 0.5x + 1
solve for x

Answer 28

so new to all of this :( Do you ever get the hang of it?

Answer 29

it gets very easy with practice

Answer 30

\[\log_{2}x+\log_{2}5=2 \]

Answer 31

indices, logs and surds one of the fun sections of math
I personally don't care for statistics i have to really pay attention then

Answer 32

lol

Answer 33

add logs multiply the numbers

Answer 34

so 2^2 = 5 x
x =

Answer 35

\[\log_{2}5x=2 \] how do i isolate the x?

Answer 36

see above

Answer 37

x=4/5

Answer 38

you got it already see how easy

Answer 39

just need to know the log laws

Answer 40

they are so simple but so easy to forget :(

Answer 41

be careful when converting bases

Answer 42

\[\log_{6}(x-2)+\log_{6}(x+3)=1 \]

Answer 43

hint log base 6 of 6 = 1

Answer 44

also I put this up before but I have to solve for x and can't remember where I wrote it \[3^{x}=30\]

Answer 45

what do you have for the first one?

Answer 46

(x-2) (x+3)=6?

Answer 47

yes keep going

Answer 48

x log base 10 of 3 = log base 10 of 30
x = log base 10 of 30/ log base 10 of 3

Answer 49

are you allowed to use the calculator?

Answer 50

yes

Answer 51

well you can do the 2nd one like that
did you get 4, -3 for the other one

Answer 52

\[x^{2}-5x=0???\]

Answer 53

oooo I am a bit lost LOL

Answer 54

(x-2) (x+3)=6
x(x + 3) -2(x + 3) = 6
x^2 + 3x -2x -6 = 6

Answer 55

agh I didn't have a plus on my 3.... so (x-4) (x+3) = 0

Answer 56

so 4, -3

Answer 57

would it be true to say that -3 is not a solution as you can't have a log of a negative?

Answer 58

i didn't check in the original. always a good practice

Answer 59

same page

Answer 60

it looks like 4 is not a solution??

Answer 61

why

Answer 62

i think I am calculating wrong ahhhh I am a numbskull :(

Answer 63

log14/log 6 =1.14/.77=1.48
log6(x−2)+log6(x+3)=1
log base 6 of (4-2)*(4 + 3) 2*7 =log base 6 of 14

Answer 64

so it doesn't equal 1?

Answer 65

log base 6 of 14 = 1.48 which is almost 1

Answer 66

ok i think lol

Answer 67

oh dear I just had a look it is (x-2) + (x+3) not multiply :(
@triciaal

Answer 68

log6(x−2)+log6(x+3)=1
when we add logs we multiply the numbers in this case expressions
? why do you say not multiply?

Answer 69

oh so it still becomes (x-2) (x+3) = 1 you must think I am a twat LOL