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x =4 is this right?
yes - i've checked it and it seems correct - this is my working
log2 (x^2) + log2 (x -1 ) = log2 (5x + 4) +1
so log2 (x^3 - x^2) = log2 (5x+4) + log2 (2)
= log 2(10x+8)
so x^3 - x^2 = 10x +8
and 0 = x^3 - x^2 -10x +8
= (x-4)(x^2 +3x+2)
= (x-4)(x+2)(x+1)
x =4, -1 or -2
but logs of negatives have no meaning, so x=4 is the only solutiom
log2 (x^2) + log2 (x -1 ) = log2 (5x + 4) +1
so log2 (x^3 - x^2) = log2 (5x+4) + log2 (2)
= log 2(10x+8)
so x^3 - x^2 = 10x +8
and 0 = x^3 - x^2 -10x +8
= (x-4)(x^2 +3x+2)
= (x-4)(x+2)(x+1)
x =4, -1 or -2
but logs of negatives have no meaning, so x=4 is the only solutiom
yep - they nearly always want you to express things in teh same base and then equate
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