# System of exponential equations that also involve logs

### System of exponential equations that also involve logs

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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.
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Learningforcomp

Posts: 9
Joined: Fri Sep 11, 2020 2:35 pm
Reputation: 0

### Re: System of exponential equations that also involve logs

$$\begin{array}{|l} x^{y-2}= 4 \\ x^{2y-3} = 64 \end{array}$$

$$\begin{array}{|l} \frac{x^{y}}{x^{2}} = 4 \\ \frac{x^{2y}}{x^{3}} = 64 \end{array}$$

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

$$\begin{array}{|l} (x^{y})^{2} = 16x^{4} \\ (x^{y})^{2} = 64x^{3} \end{array}$$

16$$x^{4}$$=64$$x^{3}$$ $$\Rightarrow$$ x=4
We return to (1).

$$4^{y}$$=4.$$4^{2}$$ ; $$4^{y}$$=$$4^{3}$$ ; y=3

(4;3)
Guest

### Re: System of exponential equations that also involve logs

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

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

HallsofIvy

Posts: 263
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 98

### Re: System of exponential equations that also involve logs

HaIIsofIvy-big mistake

$$4^{log_{16 }4}$$ .16=?$$4^{\frac{5}{2}}$$

$$4^{\frac{1}{2}}$$.$$4^{2}$$=?$$4^{\frac{5}{2}}$$

$$4^{\frac{1}{2}+\frac{4}{2}}$$=?$$4^{\frac{5}{2}}$$

$$4^{\frac{5}{2}}$$=$$4^{\frac{5}{2}}$$
Guest

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