Logarithm help...?

Question

The equation given is
lg5lg5 + lg5lgx = lg4lg4 + lg4lgx
Find x

Help please >.<

Erm...i meant it as lg5 * lg5
To the 2nd answerer, your answer is wrong...
The answer key states that x = 1/20
But I don't know how to do >.<

Whoops nvm...
The second guy is correct until the last step...

lgx = - lg20
lgx = lg20^-1

x = 1/20

Answer 1

(log 5)² + (log 5)(log x) = (log 4)² + (log 4)(log x)
(log 5) (log x) - (log 4)(log x) = (log 4)² - (log 5)²
(log 5 - log 4) (log x) = (log 4 - log 5)(log 4 + log 5)
log x = - (log 4 + log 5)
log x = - log 20
x = 20^(-1)
x = 1 / 20

Answer 2

Did you mean lg(5) * lg(5) or lg[5lg 5]?

Second answer is correct up to the one before last line

lg x = -lg 20
You can write

-lg 20 = 0 - lg 20 = lg 1 - lg 20 (since lg 1 = 0)

lg 1 - lg 20 = lg(1/20)

That's

x = 1/20

Answer 3

lg5lg5 + lg5lgx = lg4lg4 + lg4lgx
lg5lgx - lg4lgx = lg4lg4 - lg5lg5
lgx(lg5 -lg4) = lg4lg4 - lg5lg5
the RHS is a difference of two squares
lgx(lg5 - lg4) = (lg4 - lg5)(lg4 +lg5)
now (lg5 -lg4) = - (lg4 -lg5)
lgx = - (lg4 + lg5)
since lgA + lgB = lgAB
lgx = - lg20
x = - 20

Answer 4

log 5 * log 5 + log 5 log x = log 4 log 4 = log 4 log x
log 5 log x - log 4 log x = log 4 log 4 - log 5 log 5
log x ( log 5 - log 4) = ( log 4 + log 5)(log 4 - log 5)
log x = [ (log 4 + log 5)(log 4 - log 5)] / ( log 5 - log 4)
log x = [ (log 4 + log 5)(log 4 - log 5)] / - ( log 4 - log 5)
log x = - (log 4 + log 5)
log x = - log 20
log x = (log 20)^ -1
x = 1/20