SOLUTION: Solve Algebraically, approx. answer to 3 decimal places 5(10^(x-6) = 7 I have divided both sides by 5 and got 10^(x-6) = 7/5 but what are the remaining steps?

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve Algebraically, approx. answer to 3 decimal places 5(10^(x-6) = 7 I have divided both sides by 5 and got 10^(x-6) = 7/5 but what are the remaining steps?      Log On


   

Question 468838: Solve Algebraically, approx. answer to 3 decimal places
5(10^(x-6) = 7
I have divided both sides by 5 and got 10^(x-6) = 7/5 but what are the remaining steps?
Found 3 solutions by Alan3354, Earlsdon, Theo:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Solve Algebraically, approx. answer to 3 decimal places
5(10^(x-6) = 7
------------------
10%5E%28x-6%29+=+1.4
x-6 = log(1.4)
x = log(1.4) + 6
x =~ 6.146

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
5%2810%5E%28x-6%29%29+=+7 Divide both sides by 5.
10%5E%28x-6%29+=+7%2F5 Take the logarithm (common log) of both sides.
Log10%5E%28x-6%29+=+Log%287%2F5%29 Apply the power rule to the left side.
%28x-6%29Log10+=+Log%287%2F5%29 Substitute Log10+=+1
x-6+=+Log%287%2F5%29 Add 6 to both sides.
x+=+Log%287%2F5%29%2B6 Evaluate.
x+=+0.146128%2B6
highlight%28x+=+6.146%29 to three decimal places.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
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%5Ea+=+c if and only if log%28b%2C%28c%29%29+=+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%5E%28x-6%29+=+%287%2F5%29 if and only if log%2810%2C%287%2F5%29%29+=+%28x-6%29
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%2810%2C%28x%29%29 is shown as log%28%28x%29%29
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%28e%2C%28x%29%29 is shown as ln%28%28x%29%29
This would be the LN function of your calculator (assuming you are using Texas Instruments).