Log base2(x+2)+log base4(y+10)=2
Log base4(x+2)+log base2(y+10)=5/2
Solve simultaneously
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
recall log x k = (log 10 k)/(log 10 x)
or (log a k) / (log a x) , which allows me to switch bases
so log 4 (y+10) = (log 2 (y+10) ) / (log 2 4)
= (1/2) log 2 (y+10)
skipping some obvious steps:
log 2 (x+2) + (1/2)log 2 (y+10) = 2
2log 2 (x+2) + log 2 (y+10) = 4
log 2 [(x+2)^2(y+10)] = 4
(x+2)^2(y+10) = 2^4 = 16 ****
similarly
(1/2)log 2 (x+2) + log 2 (y+210) = 5/2
log 2 (x+2) + 2log 2 (y+10) = 5
log 2 [(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
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.
To solve the given system of logarithmic equations simultaneously, we'll use the properties of logarithms and change of base formula. Here are the steps to find the values of x and y:
Step 1: Convert the logarithmic equations to exponential form using the definition of logarithms.
For the first equation:
log base 2(x+2) + log base 4(y+10) = 2
We can rewrite it as:
2^2 = (x+2) * 4^(y+10)
4 = (x+2) * (2^2)^(y+10)
4 = (x+2) * 2^(2y+20)
4 = (x+2) * 2^(2(y+10))
4 = (x+2) * 2^(2y+20)
For the second equation:
log base 4(x+2) + log base 2(y+10) = 5/2
We can rewrite it as:
(5/2)^2 = (x+2) * 2^(y+10)
(5/2)^2 = (x+2) * 2^(y+10)
(5/2)^2 = (x+2) * 2^(y+10)
Step 2: Simplify the equations obtained in Step 1.
From the first equation:
4 = (x+2) * 2^(2y+20)
From the second equation:
(5/2)^2 = (x+2) * 2^(y+10)
25/4 = (x+2) * 2^(y+10)
Step 3: Equate the right-hand sides of the simplified equations.
(x+2) * 2^(2y+20) = (x+2) * 2^(y+10)
Since (x+2) is common to both sides, it can be cancelled out.
2^(2y+20) = 2^(y+10)
Step 4: Set the exponents on both sides equal to each other.
2y + 20 = y + 10
Step 5: Solve for y.
2y - y = 10 - 20
y = -10
Step 6: Substitute the value of y back into one of the original equations to find x.
Using the first equation:
4 = (x+2) * 2^(2(-10)+20)
4 = (x+2) * 2^0
4 = (x+2) * 1
4 = x + 2
x = 2
Therefore, the solutions to the system of logarithmic equations are x = 2 and y = -10.