Question

10 ^ 3x-8 =2^5-x This section is giving me major problems.

Answer 1

\[\huge 10^{3x-8}=2^{5-x}\]
here's the trick. Always make all your terms into prime numbers.

Answer 2

10 is not a prime number. So we need to split it into two numbers.

Answer 3

So we take a factor of 10 that is a prime number. What number can you think of that is a factor of 10 but also a prime number?

Answer 4

2 and 5

Answer 5

Correct.

Answer 6

Now we take a property of indices where
\[\large (ab)^2=a^2b^2\]

Answer 7

i dont understand how that helps

Answer 8

So what would the LHS look like after you've just learnt that you need to split the numbers and then this property?

Answer 9

Okay...

Answer 10

Let's split it first shall well?

Answer 11

\[\large (5\times 2)^{3x-8}=2^{5-x}\]

Answer 12

get rid of the brackets using that property I told you.

Answer 13

Let me just give you the property again without the power of 2.
\[\large (a\times b)^n=a^nb^n\]

Answer 14

5^3x-8 2^5-x

Answer 15

sorry

Answer 16

2^3x-8

Answer 17

Well Done mate. Fantastic.

Answer 18

Now this next part is like collecting like terms.

Answer 19

Have you heard about collecting like terms before?

Answer 20

i dont know how to do that

Answer 21

So you haven't heard about colelcting like terms when you did algebra.
Collecting like terms:
x+x+y+2y=2x+3y?

Answer 22

i dont have a clue with this mess

Answer 23

Okay...
\[\huge 5^{3x-8}2^{3x-8}=2^{5-x}\]
\[\huge 5^{3x-8}=\frac{2^{5-x}}{2^{3x-8}}\]

Answer 24

Simplify the RHS (Right Hand Side) for me please.

Answer 25

is it 5-x/3x-8?

Answer 26

\[\huge \frac{a^{n-1}}{a^{n+3}}=a^{(n-1)-(n+3)}\]

Answer 27

oooo

Answer 28

2^(5-x)-(3x-8)

Answer 29

Correct. Now simplify the indice.

Answer 30

2^-3-4x

Answer 31

\[5-x-(3x-8)\neq 5-x-3x-8\]
\[5-x-(3x-8)=5-x-3x--8\]
\[5-x-3x--8=5-x-3x+8\]

Answer 32

is that the final answer?

Answer 33

No, that was simlifying that indice of yours. You can't seem to understand that two minus signs become a plus sign....

Answer 34

simplifying*

Answer 35

i understand that

Answer 36

So what's the simplified form of
\[\huge 2^{(5-x)-(3x-8)}\]

Answer 37

is it 13 -4x?

Answer 38

Great!! Now you're starting to get it.

Answer 39

lol i am dumb blonde

Answer 40

Okay. Next step. We just simplified the RHS.
\[\huge 5^{3x-8}=2^{13-4x}\]

Answer 41

but i am just a sophmore

Answer 42

that means nothing to me

Answer 43

That still doesn't make the cut of doing good math. I was in a grade less than that, and I was still excelling at this stuff.

Answer 44

Okay we convert 5^3x-8 into log.

Answer 45

yikes

Answer 46

is it logbase5 (3x-8)

Answer 47

sorry I meant convert 2^13-4x into log. My apologies.

Answer 48

is it log 2 13-4x

Answer 49

Here's what you should be doing to convert it.
\[\huge a^n=b^x\]
\[\huge log_{b}a^{n}=x\]

Answer 50

i am new to this and i dont have a clue

Answer 51

There's nothing new about maths. It doesn't matter if you're new to this website or not. Being new to the website doesn't necessarily mean you're new to math...

Answer 52

I am new to logs this week, and I have the worst math teacher in the school, I do not understand logs yet, I am sorry. I am normally a very good student.

Answer 53

You're meant to know the laws of logarithms before doing this. Is the correct answer in decimal places?

Answer 54

Because this questions is seriously a messy one. I need to round off to get the answer.

Answer 55

this question*

Answer 56

you are very smart, and this is my first week with logarithms

Answer 57

i will get them, in a few days it will be easy for me, but right now i am completely lost

Answer 58

I wanted to ask whether you have the correct answer on an answer sheet? If you do is the correct answer in decimal places?

Answer 59

I do not have any answer to the problem.

Answer 60

Okay. we'll continue. What did you get when you convert the terms into log. Look at that property I gave you.

Answer 61

I dont know how to do it

Answer 62

Here's a straight example. Apply this example to our current method okay?
\[\huge x^{n+1}=y^{n-2}\]
\[\huge log_{x}y^{n-2}=n+1\]

Answer 63

current question**

Answer 64

I have 2^-4x+13

Answer 65

You make 2 the log base.

Answer 66

the log base is the n in this:
\[\huge log_{n}<---log base\]

Answer 67

log base 2

Answer 68

Yes. Now the number beside log base 2 is the WHOLE TERM on the LHS.

Answer 69

THe whole term is \[\huge 5^{3x-8}\]

Answer 70

log base 2 5^3x-8 = -4x +13

Answer 71

Good work.

Answer 72

Now we need to apply another property of logs...
\[\huge log_{n}x^a=alog_{n}x\]

Answer 73

i thought we were done

Answer 74

The a is the x^a can be moved to the side of the log. And no, we aren't done. You're meant to solve for x, aren't you?

Answer 75

the a in*

Answer 76

yes

Answer 77

wait

Answer 78

3x-8 log base 2 5

Answer 79

what? this is what you had:
log base 2 5^3x-8 = -4x +13?

Answer 80

And yes you're correct with the LHS.

Answer 81

ty

Answer 82

does the right stay the same

Answer 83

\[\huge (3x-8)log_{2}5=13-4x\]

Answer 84

Yes.

Answer 85

Move the (3x-8) to the RHS.

Answer 86

\[\huge log_{2}5=\frac{13-4x}{3x-8}\]

Answer 87

yes i have that

Answer 88

Then, we get rid of the logs.
We will get this.
\[\huge 2^{\frac{13-4x}{3x-8}}=5\]

Answer 89

to find what the indice of 2 is equal to, we must use logs again.
So...
\[\frac{13-4x}{3x-8}=\frac{log5}{log2}\]
You should be able to find what the RHS is equal to when you plug that into your calculator.

Answer 90

Thank you very much, I have to go to bed, i am up hours past my bed time

Answer 91

I will be back and hope I see you again

Answer 92

Well you can probably find x by yourself after that.

Answer 93

And no worries.

Answer 94

thank you

Answer 95

Good night.

Answer 96

you too