Question 468838
the problem is:
5 * (10^(x-6) = 7
divide both sides of the equation by 5 to get:
10^(x-6) = 7/5
you have 2 ways to solve this.
THE FIRST WAY TO SOLVE THIS IS AS FOLLOWS:
you can take the log of each side of the equation to get:
log(10^(x-6)) = log(7/5)
from the rules of logarithms, log(a^b) = b*log(a), so your equation becomes:
(x-6)*log(10) = log(7/5)
divide both sides of this equation by log(10) to get:
x-6 = log(7/5) / log(10)
solve for x-6 to get:
x-6 = .146128036
add 6 to both sides of this equation to get x = 6.146128036
confirm by replacing x in your original equation with 6.146128036 to see if the equation is true.
your original equation is:
5 * (10^(x-6) = 7
that becomes 5 * 10^.146128036 = 7 which becomes 7 - 7 so the answer is good.
THE SECOND WAY TO SOLVE THIS IS TO RECOGNIZE THAT:
b^a = c if and only if log of c to the base b is equal to a.
this looks like the following:
{{{b^a = c}}} if and only if {{{log(b,(c)) = a}}}
your original equation is:
5(10^(x-6) = 7 
divide both sides of this equation by 5 to get:
10^(x-6) = (7/5)
looking at the equation I just showed you of:
b^a = c if and only if log of c to the base b is equal to a.
b would be equal to 10
a would be equal to x-6
c would be equal to 7/5.
replace that in the equation i just showed you of:
b^a = c if and only if log of c to the base b is equal to a to get:
10^(x-6) = (7/5) if and only if log of (7/5) to the base 10 is equal to (x-6)
this looks like:
{{{10^(x-6) = (7/5)}}} if and only if {{{log(10,(7/5)) = (x-6)}}}
your equation to solve in this case is log(7/5) = x-6 which makes x-6 equal to .146128036.
YOU GET THE SAME ANSWER EITHER WAY
Note that log of x to the base 10 is just shown as log (x)
This looks like:
{{{log(10,(x))}}} is shown as {{{log((x))}}}
This would be the LOG function key of your calculator (assuming you are using Texas Instruments).
Note that log of x to the base e is just shown as ln(x)
This lookelike:
{{{log(e,(x))}}} is shown as {{{ln((x))}}}
This would be the LN function of your calculator (assuming you are using Texas Instruments).