Write both sides as powers of 16:
[tex]\log_4(2-x)=\log_{16}(13-4x) \implies 16^{\log_4(2-x)}=16^{\log_{16}(13-4x)}[/tex]
Since 16 = 4², we can rearrange the left side to get
[tex]16^{\log_4(2-x)}=(4^2)^{\log_4(2-x)}=4^{2\log_4(2-x)}=4^{\log_4(2-x)^2}[/tex]
Then, recalling that [tex]b^{\log_ba}=a[/tex], we have
[tex]4^{\log_4(2-x)^2}=16^{\log_{16}(13-4x)}\implies(2-x)^2=13-4x[/tex]
Now solve for x :
[tex](2-x)^2=13-4x[/tex]
[tex]4-4x+x^2=13-4x[/tex]
[tex]x^2-9=0[/tex]
[tex](x+3)(x-3)=0[/tex]
[tex]\implies x=-3\text{ or }x=3[/tex]
Notice that if x = 3, then log₄(2 - x) = log₄(-1) is undefined, so we throw out this solution. We don't run into this problem with x = -3, so that's the only (real) solution.
