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# Help meee

-5
1
2
+211

Solve the system of equations
y = \log_2 (2x)
y = log_4 (16 + x)

May 24, 2024

### 2+0 Answers

#1
+806
0

First off, we can subsitue y out to get $$\log _4\left(16+x\right)=\log _2\left(2x\right)$$

Now, we just have to solve for x. $$\log _4\left(16+x\right)=\log _2\left(x\right)+1$$

I'm not sure exactly how to explain, but now we just have to isolate x and we eventually get

$$x=\frac{1+\sqrt{257}}{8}$$

Here's a graph to make it more clear.

Their intersection point is our answer for x!

Thanks! :)

May 24, 2024
#2
+129495
+1

log2 (2x)  = log4 (16 + x)

Using the change-of-base theorem

log (2x) / log2  = log (16 + x) / log 4

log (2x) / log 2 = log (16 + x) / log 2^2

log (2x) / log 2  = log (16 + x) / [ 2 log 2]      multiply through by log 2

log (2x)  = log (16 + x) / 2

2log (2x) = log (16 + x)

log (2x)^2 = log (16 + x)      this implies that

(2x)^2 = 16 + x

4x^2 - x - 16  =  0

Using the Quadratic Formula  (and taking the positive value of x )

x =  [ 1 + sqrt [ 1^2 - 4 (4) (-16) ] ] / (2 *4)

x = [ 1 + sqrt (257) ] / 8

May 24, 2024