Solve for x: Log10^7= Log10^x - Log10^2?

Question

Solve for x: Log10^7= Log10^x - Log10^2

Answer 1

log 10^7 = log 10^x - log 10 ²
log 10^7 = log (10^x / 10 ²)
10^7 = 10^x / 10 ²
10^x = 10^9
x = 9

Answer 2

Let's see..Log10^7=7 so 7=Log 10^x-Log10^2. If you've got a good graphing calculator, you can type in everything to the right of the 7 into it as you see it. Pressing enter would give you 8. Is there No Solution??

Answer 3

Let's see..Log10^7=7 so 7=Log 10^x-Log10^2. If you've got a good graphing calculator, you can type in everything to the right of the 7 into it as you see it. Pressing enter would give you 8. Is there No Solution??

Answer 4

All answer are incorrect or partly ideal! The is a attempt no be counted if the equation holds. because of the fact the quotient x/x is given, the equation is defined for x=0, because of the fact the actuality, that there is a real decrease for x goint to 0 of x/x: limit_x->0 x/x=a million. So get rid of the quotient a minimum of for x unequal 0. The equation now's x+x²=x². This equation could be rearranged by making use of unique arithmetic operations. It supplies x=0. stupid or. What does is mean? The equation would not carry for x unequal 0. that's it. it may back be simplified a million+x=x or a million=0, that's erroneous. For 0 is holds interior the stable variety and interior the decrease attention: 0/0+0=a million+0 unequal 0. The equation would not be valid for any genuine numbers!