MATH help please and thank you?

Question

Solve for x to two decimal places;

7 to the exponent x+9 ( so like the x+9 is the 7's exponent:S )



log(little 6) x +log ( little 6) ( x+1) = 1

Answer 1

7 to the exponent x+9 ( so like the x+9 is the 7's exponent:S

This is incomplete. There is no "=" anywhere

log(little 6) x +log ( little 6) ( x+1) = 1
log_6 x + lof _6 (x+1)=1
log_6 (x(x+1))=1
x(x+1)=6
x^2+x-6=0
(x+3)(x-2)=0
x+3=0
x=-3
x-2=0
x=2

x=2, -3

Answer 2

The first one can't be solved because you didn't supply both sides of the equation.

For the second,
logarithms can be added together so that logf(x) + log g(x) = log(f(x)g(x))
so we get log(base 6) (x*(x+1)) = 1
Then make both sides the power of 6
so 6^log(base 6) (x*(x+1)) = 6^1 = 6
so x(x+1) = 6
x^2 + x - 6 = 0
Then factor:
(x+3)(x-2) = 0
x = -3 or x = 2

Answer 3

log[6](x^2+x)=1 (multiply the x and x+1)
6=x^2+x (exponential form
0=x^2+x-6
0=(x+3)(x-2)
x=-3,x=2

You cannot take the log of a negative number, so x=2 is your solution

(the first one does not have an = sign)

Answer 4

what do you want to do with the first eqn?
7^(x+9) = what?

Log6x + log6(x+1) = 1
is the same as log6(x(x=1))=1
raise both sides to 6
x(x+1) = 6
x^2 + x - 6 = 0
(x + 3)(x -2)=0
x = -3 and x = 2

Answer 5

log(little 6) x +log ( little 6) ( x+1) = 1
log_6 x(x+1)=1
6^(x(x+1)) = 6^1
x^2+x = 1
x^2+x-1=0
x = [-1+/- sqrt(1+4)]/2
x = -1+/- sqrt(5)
x = 1.24