Question

if 2log8+log7-2log4=log3-logx+log21 find x​

Answer 1

Answer:

Step-by-step explanation:

given 2log8+log7-2log4=log3-logx+log21

8 can be written as 2^3 and 4 can be written as 2^2

21 can be written as 3*7

now

log(a*b)=loga+logb

now log21=log3*7=log3+log7

log8=log2^3=3log2

log4=log2^2=2log2

now

2*3log2+log7-2*2log2=log3-logx+log3+log7

6log2+log7-4log2=2log3+log7-logx

2log2=2log3-logx

logx=2log3-2log2

=(log(3^2)-log(2^2))

=log9-log4

=log(9/4)[loga-logb=loga/b]

both sides equal so eliminate log on both sides

x=9/4

Answer 2

(●'◡'●)(●'◡'●)

Answer:

2log8 + log 7 - 2log4 = log3 - logx + log21

Using Property

• log(m×n) = logm + logn

• log(m^n) = n×logm

6log2 + log7 - 4log2 = log3 - logx + log3 + log7

On Simplifying

2log2 = 2log3 - logx

logx = 2log3 - 2log2

logx = 2 log(3/2)

logx = log(9/4)

So,

x = 9/4