System of exponential equations that also involve logs

[Learningforcomp]

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Greetings. I need your help solving the following attached problems:

The solutions are: for the first( the one with logs): S{4;16}

for the second:S={4;3}

also If you are available to do so any tips would be appreciated.

Your help is deeply appreciated.

Thank you.

[Guest]

[tex]\begin{array}{|l} x^{y-2}= 4 \\ x^{2y-3} = 64 \end{array}[/tex]

[tex]\begin{array}{|l} \frac{x^{y}}{x^{2}} = 4 \\ \frac{x^{2y}}{x^{3}} = 64 \end{array}[/tex]

[tex]\begin{array}{|l} (1) x^{y} = 4x^{2} (\text{We raise to the second degree.})\\ (x^{y})^{2} = 64x^{3} \end{array}[/tex]

[tex]\begin{array}{|l} (x^{y})^{2} = 16x^{4} \\ (x^{y})^{2} = 64x^{3} \end{array}[/tex]

16[tex]x^{4}[/tex]=64[tex]x^{3}[/tex] [tex]\Rightarrow[/tex] x=4
We return to (1).

[tex]4^{y}[/tex]=4.[tex]4^{2}[/tex] ; [tex]4^{y}[/tex]=[tex]4^{3}[/tex] ; y=3

(4;3)

[HallsofIvy]

[tex]x^{log_y(x)}\cdot y= x^{5/2}[/tex]
If x= 4 and y= 16 then [tex]log_y(x)= log_4(16)= 2[/tex] since [tex]4^2= 16[/tex].
So the left side is [tex]4^{2}(16)= 16(16)= 256. And [tex]4^{5/2}= 2^5= 32[/tex]!

No, x= 4, y= 16 is NOT a solution!

[Guest]

HaIIsofIvy-big mistake

[tex]4^{log_{16 }4}[/tex] .16=?[tex]4^{\frac{5}{2}}[/tex]

[tex]4^{\frac{1}{2}}[/tex].[tex]4^{2}[/tex]=?[tex]4^{\frac{5}{2}}[/tex]

[tex]4^{\frac{1}{2}+\frac{4}{2}}[/tex]=?[tex]4^{\frac{5}{2}}[/tex]

[tex]4^{\frac{5}{2}}[/tex]=[tex]4^{\frac{5}{2}}[/tex]