log10 (x − 2) − 3log10 2x = 1 − log10 y, solve for y?

All logs to base 10
log(x-2) - log((2x)³) = log(10) - log(y)
log( (x-2)/8x³ ) = log(10/y )

(x-2) / 8x³ = 10 / y
y(x-2) = 80x³
y = 80x³ / (x-2)

Let log be log to base 10

log ( x - 2 ) - log ( 2x - 1 )^3 = 1 - log y

log y = 1 - log ( x - 2 ) + log ( 2x - 1 )^3

log y = 1 + log [ ( 2x - 1 )^3 / ( x - 2 ) ]

log y = log 10 + log [ ( 2x - 1 )^3 / ( x - 2 ) ]

log y = log [ 10 ( 2x - 1 )^3 / ( x - 2 ) ]

y = 10 ( 2x - 1 )^3 / ( x - 2 )

log₁₀ (x − 2) − 3 log₁₀ 2x = 1 − log₁₀ y
log₁₀ y = 1 + 3 log₁₀ 2x − log₁₀ (x − 2)
log₁₀ y = log₁₀ 10 + log₁₀ (2x)³ − log₁₀ (x − 2)
log₁₀ y = log₁₀ [10*(2x)³ / (x - 2)]
log₁₀ y = log₁₀ [80x³ / (x - 2)]

y = 80x³ / (x - 2)

Log10 (x − 2) − 3log10 2x = 1 − log10 y
Log (x − 2) − 3 log 2x = log 10 − log y
Log (x − 2) − log [(2x)^3] = log 10 − log y
Log (x-2)/[(2x)^3] = log 10/y
(x-2)/[(2x)^3] = 10/y
y = 10[(2x)^3]/(x-2)

Since 10 is the base of the standard log anyway, I'm just going to write these as "log".
log(x-2) - 3(log(2x)) = 1 - log(y)
log(x-2) - log((2x)^3) = 1 - log(y)
log(x-2) - log(8x^3) = 1 - log(y)
log((x-2) / 8x^3) = 1 - log(y)
log((x-2) / 8x^3) - 1 = -log(y)
log((x-2) / 8x^3) - log(10) = -log(y)
log((x-2) / 80x^3) = -log(y)
log((x-2) / 80x^3) = log(y^-1)
(x-2) / 80x^3 = y^-1
y = 80x^3 / (x-2)

log10 (x − 2) − 3log10 2x = 1 − log10 y
log10 y=1-1og10 (x-2)+3log10 2x
y=10^[1-1og10 (x-2)+3log10 2x]