# Systems of Exponential and Logarithmic Equations

When we solve systems of exponential and logarithmic equations, we use standard techniques for solving exponential and logarithmic equations, and standard techniques for solving system of equations.

Example. Solve the following system of equations: {(log_2(x)+log_4(y)=4),(3^(x^2)=9*3^(15y+2)):}.

Examine first equation of the system. Since log_2(x)=log_(2^2)(x^2)=log_4(x^2) then equation can be rewritten as log_4(x^2)+log_4(y)=4 or log_4(x^2y)=4, i.e. x^2y=4^4=256.

Now examine second equation of the system. Since 9*3^(15y+2)=3^2*3^(15y+2)=3^(2+15y+2)=3^(15y+4 then equations can be rewritten as 3^(x^2)=3^(15y+4) or x^2=15y+4.

Thus, we obtained new system, equivalent to the given system: {(x^2y=256),(x^2=15y+4):}.

Now, replace x^2 in first equation, with expression for x^2 from second equation: (15y+4)y=256 or 15y^2+4y-256=0. This equation has two roots: y_1=4, y_2=-(64)/(15).

Now, use equation x^2=15y+4.

1. If y=4 then x^2=15*4+4=64, i.e. x^2=64. This equation has two roots: x_1=8, x_2=-8.
2. If y=-64/15 then x^2=15*(-64)/15+4=-60, i.e. x^2=-60. This equation doesn't have roots.

So, we found two pairs of values: x_1=8,y_1=4 and x_2=-8,y_2=4.

Since initial system contains expressions log_2(x),log_4(y) then must satisfy following conditions: x>0,y>0. Therefore, second pair of values doesn't satisfy initial system (because x=-8<0).

That's why initial system has only one solution: (8;4).