# Hi! could you please solve for x this equation? `(log8+logx)/log32=log2/(log8-logx)` Thanks

steveschoen | College Teacher | (Level 1) Associate Educator

Posted on

Hi, thezy,

For this one, we could cross multiply first, giving:

(log 8 + log x)(log 8 - log x) = log 32 log 2

This simplifies to

(log 8)^2 - (log x)^2 = log 32 log 2

Then, we could subtract log 32 log 2 from each side, and add (log x)^2 to each side, giving us:

(log 8)^2 + log 32 log 2 = (log x)^2

Taking the square root of each side:

[ (log 8)^2 + log 32 log 2 ] = log x

Now, from here, we can write the log 8 = log 2^3 and log 32 = log 2^5.  So:

[ (log 2^3)^2 + (log 2^5) log 2 ] = log x

Simplifying:

[ (3 log 2)^2 + (5 log 2) log 2 ] = log x

9(log 2)^2 + 5(log 2)^2 = log x

14 (log 2)^2 = log x

Taking the antilog of each side:

10^[14 (log 2)^2] = x

The left side may be able to be simplified more.  I can't figure it out anymore than that, though.  Sorry.

Good luck, Thezy.  I hope this helps.

Till Then,

Steve

steveschoen | College Teacher | (Level 1) Associate Educator

Posted on

Opps.  Sorry.  But, subtract 5(log2)log2 would give us 9-5 which if 4.  So, we would have 4 instead of 14 in my final answer.

II hope that is understandable, Thezy.  Sorry.

Till Then,

Steve

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

`(log8+logx)/(log32)=log2/(log8-logx)`

`rArr (log8+logx)(log8-logx)=log2log32 `

`rArr (log8)^2-(logx)^2=log2log32 `

`rArr 9(log2)^2-(logx)^2=5(log2)^2 `

`rArr logx^2=9(log2)^2-5(log2)^2 `

`rArr logx^2=4(log2)^2 `

`rArr logx=+-(2log2)`

for `logx=log4 rArr x=4 `

for `logx=-log4 rArr x=-4`

Therefore, x=-4, 4.

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

We have

`(log(8)+log(x))/(log(32))=log(2)/(log(8)-log(x))`
Cross multiply and simplify
`(log(8)+log(x))(log(8)-log(x))=log(2)log(32)`
`(log(8))^2-(log(x))^2=log(2)log(2^5)`
`(log(2^3))^2-(log(x))^2=5(log(2))^2`
`(3log(2))^2-(log(x))^2=5(log(2))^2`
`9(log(2))^2-5(log(2))^2=(log(x))^2`
`(log(x))^2=4(log(2))^2`
`log(x)=+-2log(2)`
If
`log(x)=2log(2)`
`=>log(x)=log(2^2)`
`=>log(x)=log(4)`
`=>log(x)-log(4)=0`
`=>log(x/4)=0`
`=>x/4=1`
`=>x=4`
If
`log(x)=-2log(2)`
`=>log(x)=log(2^(-2))`
`=>log(x)=log(1/4)`
`=>log(x)-log(1/4)=0`
`=>log(x/(1/4))=0`
`=>log(4x)=0`
`=>4x=1`
`=>x=1/4`
Thus  we have  `x=4,1/4`
since
`log(8)-log(4)!=0`  and
`log(8)-log(1/4)!=0`
Therefore
both values of x are correct.