Evaluate the expression.

There is a way to get an exact answer, but it's not obvious, at first......

Note that we can write 81 as 3⁴ and we can write 256 as 4⁴

And using a log property, we can write

log₁₂ 81 + log₁₂ 256      .....as......

log₁₂ 3⁴ + log ₁₂ 4⁴ =  log₁₂(3⁴ * 4⁴) = log₁₂(3*4)⁴ = log₁₂ (12)⁴ = 4log₁₂(12) = 4*1 = 4

And that's it.......

log ₁₂  81 + log ₁₂  256

$$\frac{log81}{log12}+\frac{log256}{log12}\\\\
=\frac{log81+log256}{log12}\\\\$$

$${\frac{\left({log}_{10}\left({\mathtt{81}}\right){\mathtt{\,\small\textbf+\,}}{log}_{10}\left({\mathtt{256}}\right)\right)}{{log}_{10}\left({\mathtt{12}}\right)}} = {\mathtt{3.999\: \!999\: \!999\: \!999\: \!999\: \!7}}$$

I suspect that the exact answer is 4.  So there is probably an exact way to do this.

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I'll think about it.  

That is correct. 

Thank you Sally1    (I'll take your comment as a thank you )

I meant that I had the answer but I wanted to see how it was supposed to be worked out. 

Yes , that is what I thought Sally.

That is good - It is nice to know that my answers have been checked - we can all make mistakes.

Thank you Chris.  That's great.