SOLUTION: 15. (a) Given log10 x = 3 and log10 y = -2. show that xy - 10000y^2 = 9. (b)Solve the equations. (i) 5^(x+2) = 120+5^x (ii)log3 x = log9 (6x+7)

Question 881825: 15. (a) Given log10 x = 3 and log10 y = -2. show that xy - 10000y^2 = 9.
(b)Solve the equations.
(i) 5^(x+2) = 120+5^x
(ii)log3 x = log9 (6x+7)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
start with:
xy - 10000y^2 = 9
since 10000 is the same as 10^4, this can be rewritten as:
xy - 10^4 * y^2 = 9
you are given that log(x) = 3 and you are given that log(y) = -2
using the basic definition of logs, you can derive the following from this:
log10(x) = 3 if and only if 10^3 = x
log10(y) = -2 if and only if 10^-2 = y
you can now replace x in your original equation with 10^3 and y in your original equation with 10^-2.
you will get:
xy - 10^4 * y^2 = 9 becomes:
10^3 * 10^-2 - 10^4 * (10^-2)^2 = 9
since 10^3 * 10^-2 = 10^1 and (10^-2)^2 = 10^(-4), you get:
10^1 - 10^4 * 10^-4 = 9
since 10^4 * 10^-4 = 10^0, you get:
10^1 - 10^0 = 9 which becomes:
10 - 1 = 9 which becomes:
9 = 9

solve for 5^(x+2) = 120 + 5^x
5^(x+2) is the same as 5^x * 5^2
your equation becomes:
5^x * 5^2 = 120 + 5^x
subtract 5^x from both sides of this equation to get:
5^x * 5^2 - 5^x = 120
factor out 5^x on the left side of the equation to get:
5^x * (5^2 - 1) = 120
simplify to get:
5^x * (24) = 120
divide both sides of this equation by 24 to get:
5^x = 120/24 which becomes 5^x = 5
if 5^x = 5, this means that x has to be equal to 1 because 5^1 is equal to 5.
you could also solve this by taking the log of both sides of the equation to get log10(5^x) = log10(5) which becomes:
x * log10(5) = log10(5) which becomes:
x = 1 after you divide both sides of the equation by log10(5).

solve for log3(x) = log9(6x+7)
log3(x) = y if and only if 3^y = x
log9(6x+7) = y if and only if 9^y = 6x+7
9^y is the same as (3^2)^y which is the same as 3^(2y) which is the same as (3^y)^2
you get:
3^y = x and (3^y)^2 = 6x+7
if you square both sides of 3^y = x, you get (3^y)^2 = x^2
now you have:
(3^y)^2 = x^2 and (3^y)^2 = 6x + 7
this means that x^2 = 6x + 7
subtract 6x + 7 from both sides of this equation and you get:
x^2 - 6x - 7 = 0
factor this equation to get:
(x-7) * (x+1) = 0
solve for x to get x = 7 and x = -1
those should be your solutions.
confirm by replacing x in your original equation with 7 and then with -1 to see if the original equations are true.
when x = 7, the original equation of log3(x) = log9(6x+7) becomes:
log3(7) = log9(6*7+7) which becomes:
log3(7) = log9(49)
in order to solve these, you can convert them to exponential form as we did above, or you can use the log base conversion formula to get:
log10(7)/log10(3) = log10(49)/log10(9) and use your calculator to get:
1.771243749 = 1.771243749, confirming the solution of x = 7 is good.
you can do the same with x = -1.
i'll confirm that one using the exponential form rather than the log base conversion formula.
your original equation is log3(x) = log9(6x+7)
when x = -1, you get:
log3(-1) = log9(1)
you can stop right there, because you can't take the log of a negative number if you are looking for a real solution.
x = -1 is not a real solution to the original equation.
the only real solution is x = 7.

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