Asked by Rhonick

Log base2(x+2)+log base4(y+10)=2
Log base4(x+2)+log base2(y+10)=5/2
Solve simultaneously
Answered by Steve
so call the logs base 4 a and b.
Now, log_2(x) = 2*log_4(x) since 4=2^2
Now the system becomes

2a+b = 2
a+2b = 5/2
or, 2a+4b=5

Now subtract and you have 3b=3, so b=1
Use that to get a=1/2

so, log_4(x+2)=1/2 --> x+2=2 --> x=0
log_4(y+10)=1 --> y+10=4 --> y = -6
Answered by Reiny
recall log <sub>x</sub> k = (log <sub>10</sub> k)/(log <sub>10</sub> x)
or (log <sub>a</sub> k) / (log <sub>a</sub> x) , which allows me to switch bases

so log <sub>4</sub> (y+10) = (log <sub>2</sub> (y+10) ) / (log <sub>2</sub> 4)
= (1/2) log <sub>2</sub> (y+10)

skipping some obvious steps:
log <sub>2</sub> (x+2) + (1/2)log <sub>2</sub> (y+10) = 2
2log <sub>2</sub> (x+2) + log <sub>2</sub> (y+10) = 4
log <sub>2</sub> [(x+2)^2(y+10)] = 4
(x+2)^2(y+10) = 2^4 = 16 ****

similarly
(1/2)log <sub>2</sub> (x+2) + log <sub>2</sub> (y+210) = 5/2
log <sub>2</sub> (x+2) + 2log <sub>2</sub> (y+10) = 5
log <sub>2</sub> [(x+2)(y+10)^2] = 5
(x+2)(y+10)^2 = 2^5 = 32 ***

divide *** by ****
(y+10)/(x +2) = 5
y+10 = 5(x+2)

back in ****
(x+2)^2(y+10) = 16
(x+2)^2 (5(x+2)) = 16
(x+2)^3 = 16/5
I get x = -.52638... and back in y+10 = 5(x+2)
I get y = -2.631937...

and they work if I sub back into the original !!!
Wolfram confirmation:
http://www.wolframalpha.com/input/?i=solve+y%2B10+%3D+5(x%2B2),+(x%2B2)%5E2*(y%2B10)+%3D+16
Answered by arggghhh - Reiny
In all that weird <html>, I must have made an error.
My answer works in the first equation but not in the 2nd.

go with Steve's simple solution, I didn't see the forest for the trees.
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