log₂₀₁₁(2010x) = log₂₀₁₀(2011x)?

Let log₂₀₁₁(2010x) = log₂₀₁₀(2011x) = ⁿ. we have

2011ⁿ = 2010x -------------------- Eqn 1 and

2010ⁿ = 2011x -------------------- Eqn 2; dividing Eqn 1 by Eqn 2 we get

(2011/2010)ⁿ = (2010/2011) or

[{(2011/2010)ⁿ}/(2010/2011)] = 1 or

(2011/2010)ⁿ⁺¹ = 1 or

ⁿ⁺¹ = ⁰ or

ⁿ = ⁻¹ ----------------------------------------- Equation 3

Now multiplying Eqn 1 and Eqn 2, we get

(2011*2010)ⁿ = (2010*2011)*x² or

x² = (2011*2010)ⁿ⁻¹ = (2011*2010)⁻² or

x = 1/(2010*2011)

[As we know that e^(ln a) = a. I get the same answer as Iggy Rocko]

log(base 2011)(2010x) = log(base 2010)(2011x)

ln(2010x)/ln2011 = ln(2011x)/ln2010

(ln2010 + lnx)/ln2011 = (ln2011 + lnx)/ln2010

ln2010/ln2011 + lnx/ln2011 = ln2011/ln2010 + lnx/ln2010

lnx/ln2011 - lnx/ln2010 = ln2011/ln2010 - ln2010/ln2011

ln2011ln2010[ lnx/ln2011 - lnx/ln2010 = ln2011/ln2010 - ln2010/ln2011 ]

ln2010lnx - ln2011lnx = (ln2011)^2 - (ln2010)^2

lnx(ln2010 - ln2011) = (ln2011)^2 - (ln2010)^2

lnx = (ln2011 + ln2010)(ln2011 - ln2010)/(ln2010 - ln2011)

lnx = -ln2011 - ln2010

x = e^(-ln2011 - ln2010)