Question

find value of x.........
4^x-3^[x-(1/2)] = 3^[x+(1/2)]-2^(2x-1)

Answer 1

3^(x-1/2)=a
4^x=b
b-a=3a+1/2b
4a=b/2
b=8a......

Answer 2

demonstrate more

Answer 3

3^(x-1/2)=a 4^x=b b-a=3a+1/2b 4a=b/2 b=8a......

Answer 4

how 3^[x+(1/2)]= 3a ??

Answer 5

& 2^[2x-1]=1/2 b ??

Answer 6

As you see in @mahmit2012 's post,
Let \(3^{(x-\frac{1}{2})}=a\)
\[3^{(x+\frac{1}{2})}= 3^{(x-\frac{1}{2})+1}=3^{(x-\frac{1}{2}) \times 3} = a \times 3 = 3a\]

Answer 7

And let \(4^x=b\),
That is \(2^{2x} = b\)
So,
\[2^{2x-1}=2^{2x}\times \frac{1}{2} = b \times \frac{1}{2} = \frac{b}{2}\]
You can get these using laws of indices.

Answer 8

well \[4^x = 2^{2x}\]

slight error in the notation

\[3 \times 3^{x - \frac{1}{2}} = 3^{x - \frac{1}{2} + 1} = 3^{ x + \frac{1}{2}}\]

Answer 9

thanks to callisto,mahmit2012 & all

Answer 10

Sorry.. Something's wrong with my first post here.
It should be \[3^{(x+\frac{1}{2})}= 3^{(x-\frac{1}{2})+1}=3^{(x-\frac{1}{2})} \times 3 = a \times 3 = 3a\]

Answer 11

3(x+12)=3(x−12)+1=3(x−12)×3=a×3=3a

Answer 12

but how to solve for x ??

Answer 13

you get b=8a right now, right?

Answer 14

??

Answer 15

Do you know how to get b=8a?

Answer 16

If you do, I'll go to the next step; if not, I'll have to explain it!

Answer 17

yes, i know

Answer 18

Okay, so, now we get b=8a.
Put \( a = 3^{(x-\frac{1}{2})}\) and \(4^x=b\) into the equation. What do you get?

Answer 19

b=8a
4

Answer 20

b=8a
4^x= 8*3[x-(1/2)]

Answer 21

Yes.
Now, take log on both sides. Have you learnt logarithm yet?

Answer 22

yes,i am to equal the base like [2^x=2^3,so x=3]

Answer 23

yes,but i am trying to equal the base like [2^x=2^3,so x=3]

Answer 24

Hmm..Hope you know how it works..
When you solve 2^x=2^3, since they have the same base, you can equate the power (as it must be equal when they share the same base)
But, say, you have 5^x = 2^3, you can use log to solve x as the following:
5^x = 2^3
Take log on both sides
ln (5^x) =ln (2^3)
By log property: ln (a^b) = b ln (a), so you'll get:
x ln5 = ln 2^3 (note that 2^3 = 8)
Divide both sides by ln5 to isolate x:
x = (ln8) / (ln5)

Got it?

Answer 25

yes

Answer 26

so how to solve 4^x= 8*3[x-(1/2)] by log

Answer 27

Take log for both sides like I did.

Answer 28

log 2^2x =log 2^3*3^[x-(1/2)]

Answer 29

Not really.. Instead:
\[\ln (4^x) = \ln (8\times 3^{x-\frac{1}{2}})\]Got it so far?

Answer 30

and other step

Answer 31

Do you understand that?

Answer 32

yes i understand but not able to solve further

Answer 33

Okay, after you get that step, you have to apply the log. properties:
(1) \(\ln a^b = b\ln a\)
(2) \(\ln ab = \ln a + \ln b \)
Any problem so far?

Answer 34

i know all the rule of log

Answer 35

Alright, so it's all about application here.
\[\ln (4^x) = \ln (8\times 3^{x-\frac{1}{2}})\]\[\ln (4^x) = \ln (8) + \ln (3^{x-\frac{1}{2}})\]Any problems?

Answer 36

yes, no problem

Answer 37

Apply the log. property (2).
\[x\ln (4) = \ln (8) + ({x-\frac{1}{2}})\ln (3)\]
Any questions?

Answer 38

yes, no problem

Answer 39

\[x\ln (4) = \ln (8) + {x \ln3 -(\frac{1}{2}})\ln (3)\]
Got it?

Answer 40

yes

Answer 41

Now, solve the equation by simplifying the constant term and doing the addition/subtraction for like terms. Can you do it?

Answer 42

sorry, more..

Answer 43

One more hint:
Subtract xln3 from both sides:
\[x\ln (4) - x \ln3= \ln (8) -(\frac{1}{2})\ln (3)+ x \ln3 - x \ln3\]\[x(\ln (4) - \ln3)= \ln (8) -(\frac{1}{2})\ln (3)\]Can you do it now?

Answer 44

@wwe123 Where are you stuck at?

Answer 45

in find the value of x
x(ln(4)−ln3)=ln(8)−(12)ln(3)

Answer 46

it not 12 ,1/2

Answer 47

The rest is just sum manipulations. Can you tell me exactly where you don't know how to do?

Answer 48

(4^x)/3 =8/(3½)

Answer 49

How do you get that?

Answer 50

for got that
plz solve
x(ln(4)−ln3)=ln(8)−(1/2)ln(3)

Answer 51

How would you solve this equation?

Answer 52

no

Answer 53

Have you tried to solve it?

Answer 54

yes, a have a qurry this question is of quadratic equation

Answer 55

quadratic?
Can you show me how you solve it?

Answer 56

no i am not able solve it . neither by quadratic equation nor by log

Answer 57

just show me your attempt..

Answer 58

i try your method but not able to solve it by quadratic equation

Answer 59

it is quadratic equation, how we take 2 variable & solve it by log

Answer 60

Why is it a quadratic equation?

Answer 61

There is only one variable in this question, that is x.

Answer 62

solve this ques ans is x=(3/2)

Answer 63

how do you get this? Can you show your workings?

Answer 64

i know ans only becoz its give in my math book

Answer 65

((((quadratic equation)))))

Answer 66

Phew. We did something wrong!

Answer 67

Let's start all over again!
The question is \(4^x-3^{(x-\frac{1}{2})} = 3^{(x+\frac{1}{2})}-2^{(2x-1)}\)
Let \(a =3^{(x-\frac{1}{2})}\) and \(b=4^x=2^{2x}\)
\[3^{(x+\frac{1}{2})} = 3^{(x-\frac{1}{2})+1}= 3^{(x-\frac{1}{2})} \times 3 =3a\]\[2^{(2x-1)}= \frac{2^{2x}}{2}=\frac{b}{2}\]

\[4^x-3^{(x-\frac{1}{2})} = 3^{(x+\frac{1}{2})}-2^{(2x-1)}\]\[b - a = 3a - \frac{b}{2}\]\[\frac{3b}{2} = 4a\]\[b =\frac{ 8a}{3}\]

Answer 68

Now, put \(a =3^{(x-\frac{1}{2})}\) and \(b=4^x\) into the equation.
\[4^x = \frac{8}{3} (3^{(x-\frac{1}{2})})\]Take log. for both sides.\[\ln 4^x = \ln [\frac{8}{3} (3^{(x-\frac{1}{2})})]\]Use log property log(ab) = log (a) + log (b) for the right side.\[\ln 4^x = \ln (\frac{8}{3}) + \ln (3^{(x-\frac{1}{2})})\]Apply the log property log (a^b) = b log(a) for both sides.\[x \ln 4 = \ln (\frac{8}{3}) + (x-\frac{1}{2})\ln (3)\]\[x \ln 4 = \ln (\frac{8}{3}) + x\ln (3)-\frac{1}{2}\ln (3)\]

Answer 69

wait

Answer 70

euraka ,i done it

Answer 71

do not use log

Answer 72

its quadratic equation

Answer 73

see this

Answer 74

b=8a/3

Answer 75

4^x = (8/3)*3^[x-(1/2)]
2^2x = 8^[x-(1/2)]
2^2x = 2^3[x-(1/2)]
2x=3[x-(1/2)]
2x=3x-(3/2)
2x=(6x-3)/2
4x=6x-3
2x-3=0
x=3/2
and answer is (3/2)

Answer 76

Sorry, would you mind explaining the step 2^2x = 8^[x-(1/2)], how do you get this?

Answer 77

\[2^{2x} = \frac{8}{3} \times 3^{(x-\frac{1}{2})}\]\[2^{2x} = 8\times 3^{(x-\frac{1}{2})-1}\]\[2^{2x} = 2^3\times 3^{(x-\frac{3}{2})}\]Divide both sides by 2^3
\[\frac{2^{2x}}{2^3} = \frac{2^3\times 3^{(x-\frac{3}{2})}}{2^3}\]\[2^{2x-3} = 3^{(x-\frac{3}{2})}\] You still need to take log to solve it!