solve log problem step by step please?

Question

if 8^(2-x)=4^(3x) then x=

Answer 1

There are two different ways of doing this. First, you could take the log of both sides, then isolate x:

log (8^(2-x)) = log (4^(3x))
(2-x) log (8) = (3x) log (4)
2 log(8) - x log (8) = (3x) log (4)
2 log(8) = (3x) log (4) + x log (8)
2 log(8) = x (3 log (4) + log (8))
x = 2 log(8) / (3 log (4) + log (8))

From here, you can use more log properties to simplify:
x = 2 / (3 log(4)/log(8) + 1)
x = 2 / (3 log(2^2)/log(2^3) + 1)
x = 2 / (3*2log(2)/3log(2) + 1)
x = 2 / (3*2/3 + 1)
x = 2 / (2 + 1)
x = 2/3

Or in the second step, do this:
(2-x) log (8) = (3x) log (4)
(2-x) log (8)/log (4) = (3x)
(2-x) log (2^3)/log (2^2) = (3x)
(2-x) (3/2) = (3x) etc.

You could also do this:
8^(2-x)=4^(3x)
(2^3)^(2-x)=(2^2)^(3x)
(2)^3(2-x)=(2)^2(3x)
3(2-x) = 2(3x)
6 - 3x = 6x
6 = 9x
x = 2/3

Answer 2

Given,

8^(2 - x) = 4^(3x)

Log both sides,

log [8^(2 - x)] = log [4^(3x)]

(2 - x) log 8 = 3x log 4

2 log 8 - x log 8 = 3x log 4

2 log 8 = 3x log 4 + x log 8

2 log 8 = x (3 log 4 + log 8)

x = (2 log 8)/(3 log 4 + log 8)

x = (2 log 2^3)/(3 log 2^2 + log 2^3)

x = (6 log 2)/(6 log 2 + 3 log 2)

x = 6 log 2/[(6 + 3)(log 2)]

x = 6/9

x = 2/3

Answer 3

I think you'll get other answers from people who'll use logs. I want to show that logs really aren't necessary for this particular problem.

Fortunately, your two bases, 8 and 4, can both be rewritten as powers of the same base, 2.

8 ^(2-x)=4 ^(3x)
(2^3) ^(2-x) = (2^2) ^(3x)

now apply properties of exponents:
2 ^(3 * (2-x)) = 2 ^(2 * (3x))
2 ^(6 - 3x) = 2 ^(6x)

Once you have the same base on both sides, they "cancel". (actually, we're taking the log-base-2 of both sides). The point is, with the same base on both sides, all you have to do is set the powers equal to each other.
6 - 3x = 6x

which is easily solved.
:-)

Answer 4

Take log base2 of both sides:-
(2 - x) log 8 = 3x log 4
(2 - x) (3) = (3x) (2)
6 - 3x = 6x
9x = 6
x = 2 / 3

Answer 5

8^(2-x)=4^(3x)

Take the log of both sides:
ln 8^(2-x) = ln 4^(3x)

The log of a power is the eponent times the log of the base, or ln a^b = b·ln a:

(2-x)·ln 8 = (3x)·ln 4

Distribute:
2·ln 8 - x·ln 8 = 3x·ln 4

Add x·ln 8:
2·ln 8 = 3x·ln 4 + x·ln 8

Factor out x:
2·ln 8 = x(3·ln 4 + ln 8)

Divide:

x = (2 ln 8)/(3·ln 4 + ln 8)
x = (2 ln 2^3)/(3·ln 2^2 + ln 2^3)
x = (2·3·ln 2)/(3·2·ln 2 + 3·ln 2)
x = (6·ln 2)/(6·ln 2 + 3·ln 2)
x = (6·ln 2)/(9·ln 2)
x = 6/9
x = 2/3

Answer 6

8^(2 - x) = 4^(3x)
2^(6 - 3x) = 2^(6x)
6 - 3x= 6x
9x = 6
x = 2/3

There is actually no need to use logarithms to get the solution.

Answer 7

8=2^3

4=2^2

so

8^(2-x) =2^3 * (2-x)

8^(2-x) = 2^6-3x

same setp is in 4^(3x)

4^(3x) = 2^2*(3x)

4^(3x) =2^6x

2^6-3x=2^6x

here base are same

it means base are 2


IMPORTANT
when base is same

6-3x=6x

6=6X+3X

6=9X
X=6/9

X=2/3

Answer 8

with the intention to sparkling up a log project bear in mind those policies: A logarithm in condensed style has in simple terms one ln or log on the initiating, and wouldn't have any + or - in it. An greater project includes + and -, yet not multiplying or dividing. This project is in greater style. additionally bear in mind that the selection in front of the ln (organic log) is the exponent. factor your expression while needed

Answer 9

log[8^(2 -x)=4^3x]
2log8 -xlog8=3xlog4
3xlog4+xlog8=2log8
x(3log4+log8)= 2log8
x= 2log8/(3log4+log8)
x= 0.6666666666666666

Answer 10

8^(2-x)=4^(3x)
(2-x)log8=3x(log4)
(2-x)*0.903=3x*0.602
1.806-0.903x=1.806x
1.806x+0.903x=1.806
2.709x=1.806
x=1.806/2.709= 0.666666666