Question

4^(3x-1) = 3^(x-2)

Answer 1

x is not going to be an integer...

Answer 2

take logs to both sides... can you do it?

Answer 3

3x-1 log 4 = x-2 log 3?

Answer 4

just use brackets ... or you can take log to the base four and have

3x-1=(x-2)log_4(3)

Answer 5

then divide both sides by x-2 ...

Answer 6

....
Taking log both sides
3xlog4+2log3=xlog3+log4
x(3log4-log3) <---- here .. how come x(3log4-log3)
x(log64/3)=2log2/3

Answer 7

@sahil51993

Answer 8

what?????

Answer 9

...
Taking log both sides
3xlog4+2log3=xlog3+log4




x(3log4-log3) <---- here .. why x(3log4-log3)

Answer 10

this can be written as 4^3x*3^2=3^x*4 Taking log both sides 3xlog4+2log3=xlog3+log4 x(3log4-log3)=log4-2log3 x(log64/3)=2log2/3 x=2(log2/3)/6(log2/3) x=2/6=1/3

Answer 11

the domain of x should be given.

Answer 12

you're still doing the same step which one I'm asking you to explain.....

sahil51993
Medals 0

this can be written as 4^3x*3^2=3^x*4 Taking log both sides 3xlog4+2log3=xlog3+log4 x(3log4-log3)=log4-2log3 x(log64/3)=2log2/3 x=2(log2/3)/6(log2/3) x=2/6=1/3


from here...
3xlog4+2log3=xlog3+log4

how does this came?
x(3log4-log3)=log4-2log3

Answer 13

leve it... I see sorry.

Answer 14

simply bu moving term containing x to left and others to right

Answer 15

the problem is that you passed steps while solving ... and the way of typing got me confused that's why I asked...

Answer 16

I'm stuck here:
3x-1/x-2 = x-2 log4(3)/x-2
Is this right?

Answer 17

@math102 if you follow steps that @sahil51993 just provided... you can easily solve this without calculator. :)

Answer 18

Hey lad u knw this property
a^(c-d) =(a^c)/(a^d)
do this and then apply log as i mentioned earlier

Answer 19

@math102 can you follow the steps that sahil wrote?

Answer 20

So the first part looks like this? \[4^{3x}\times3^{2}=3^{x}\times4\]

Answer 21

([4^{3x})*times3^{2}=(3^{x})*times4

Answer 22

How I have it above correct?

Answer 23

yes. it's correct...

Answer 24

Then:
\[3x \log4+2 \log 3= \log4x (3\log4-\log3) ?\]

Answer 25

Sorry, i'm confused how its written above... i'm trying to follow it step by step.

Answer 26

@sahil51993 I think you made a mistake in your solution... it's not 64/3 it's 3/64

Answer 27

\[ \LARGE \begin{array}{l}{4^{3x - 1}} = {3^{x - 2}}\\\frac{{{4^{3x}}}}{4} = \frac{{{3^x}}}{{{3^2}}}\\\\{2^2} \cdot {3^x} = {3^2} \cdot {4^{3x}}\\\log \left( {{2^2} \cdot {3^x}} \right) = \log \left( {{3^2} \cdot {4^{3x}}} \right)\\\\\log \left( {{2^2}} \right) + \log \left( {{3^x}} \right) = \log \left( {{3^2}} \right) + \log \left( {{4^{3x}}} \right)\\\\2\log \left( 2 \right) + x\log \left( 3 \right) = 2\log \left( 3 \right) + 3x\log \left( 4 \right)\\\\x\log \left( 3 \right) - 3x\log \left( 4 \right) = 2\log \left( 3 \right) - 2\log \left( 2 \right)\\\\x\left[ {\log \left( 3 \right) - 3\log \left( 4 \right)} \right] = 2\left[ {\log \left( 3 \right) - \log \left( 2 \right)} \right]\\\\x\left[ {\log \left( 3 \right) - \log \left( {{4^3}} \right)} \right] = 2\left[ {\log \left( {\frac{3}{2}} \right)} \right]\\\\x\left[ {\log \left( {\frac{3}{{{4^3}}}} \right)} \right] = 2\left[ {\log \left( {\frac{3}{2}} \right)} \right]\\\\x\left[ {\log \left( {\frac{3}{{{2^6}}}} \right)} \right] = 2\left[ {\log \left( {\frac{3}{2}} \right)} \right]\\x = \frac{{2\left[ {\log \left( {\frac{3}{2}} \right)} \right]}}{{\left[ {\log \left( {\frac{3}{{{2^6}}}} \right)} \right]}}\\\\wolframa\\\\x \approx - 0.26\\\\MY\,PC\\\\x \approx - 0.13\end{array}\]

Answer 28

this can be written as 4^3x*3^2=3^x*4
Taking log both sides
3xlog4+2log3=xlog3+log4
x(3log4-log3)=log4-2log3
x(log64/3)=2log2/3
x=2(log2/3)/6(log2/3)
x=2/6=1/3

Answer 29

This is correct as far as I am concerned

Answer 30

well check this out then... http://www.wolframalpha.com/input/?i=solve+4^%283x-1%29+%3D+3^%28x-2%29

Answer 31

@sahil51993

Answer 32

On the very first step how did you get 4^3x4/4 where did you get the 4 in the denominator?

Answer 33

\[\LARGE 4^{3x-1}=something \]
\[\LARGE 4^{3x}\times 4^{-1}=something \]
\[\LARGE 4^{3x}\times \frac14 =something \]
\[\LARGE \frac{4^{3x}}{4} =something \]

Answer 34

there is four in denominator on lhs and 9 in denominator in RHS

Answer 35

I don't know where to you mean..... which row , where?

Answer 36

(4^3x)/4=3^x/9
this is our 1st step
u knw this property a^(c-d) =(a^c)/(a^d)

Answer 37

@sahil51993 As much as I know \(\LARGE 3^2=9\)

Answer 38

I don't know where do you think it's wrong.... actually our "fisrst" steps are the same... o.O

Answer 39

Okay I got -.2650 if I round to the 4 decimal spaces. Do you both agree this is correct?

Answer 40

I also see you got -.13 how did you get that answer?

Answer 41

I used calculator in my PC , but I think wolframa's right... it never fails.... If I were in your position , I'd ignore -0.13 . Now It's up to you which one you'll take although I'm 70% sure that -0.26 is "more" correct. I did what I had to do.... now you're up.

Answer 42

Okay, thank you both for your help.

Answer 43

yw