Question

Solve for x, where x is a real number.

1) log x+ log(x-3) =1
2) |x-3|< 7

Could any of u guys please explain to me how to solve this two step by step, i am so rusty on my math.thanks so much

Answer 1

in 1... u can apply laws of log m + log n = log mn.... then get inverse log both side to get rid of log... then you can solve for x...

Answer 2

in 2... solve for x first and get its absolute value and test for inequality

Answer 3

so i get logx(x-3)=1 and then what? @Orion1213

Answer 4

inv log both side.... inv log LHS = inv log RHS, this will cancel log... inv log a = 10^a

Answer 5

so it would be equal to log 10, because 10 is the base right?

I'll have
log x^2-3x = log 10
x^2-3x=10
x^2-3x-10 & then just factor

Answer 6

@Orion1213 ^

Answer 7

wrong format for inv log... it should be

10^[log x(x-3)] = 10^1

btw well get the same... yeah go on factoring...

Answer 8

what is the factor of -10 which will give you a sum of -3?

Answer 9

so after i do x^2-3x-10=0 we'll have (x-5)(x+2)=0
The answer is x=5, because the answer cant be negative right?@Orion1213

Answer 10

yes you're correct.... \(\ddot\smile\)