SOLUTION: Solve log6(3.5) + log6(2) A 1.09 B 1.32 C 1.43 D 1.45 E 6.32 F 7.09 I tried what I thought was the correct way of solving this,but i came to an answer which isnt a choi

Algebra ->  Algebra  -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve log6(3.5) + log6(2) A 1.09 B 1.32 C 1.43 D 1.45 E 6.32 F 7.09 I tried what I thought was the correct way of solving this,but i came to an answer which isnt a choi      Log On

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Question 124443: Solve log6(3.5) + log6(2)
A 1.09
B 1.32
C 1.43
D 1.45
E 6.32
F 7.09
I tried what I thought was the correct way of solving this,but i came to an answer which isnt a choice,so im not sure what to try.
If someone could show me the steps to come to the correct answer it would be very much appriciated. Thank you very much for your time!

Found 2 solutions by ankor@dixie-net.com, bucky:
Answer by ankor@dixie-net.com(15645) About Me  (Show Source):
You can put this solution on YOUR website!
Solve log6(3.5) + log6(2):
:
Solve for y:
log6(3.5) + log6(2) = y
:
log6(3.5*2) = y; adding logs means multiply
:
log6(7) = y
:
6%5Ey+=+7; exponent equiv of logs
Find the common log of both sides
log(6^y) = log(7)
:
Use the log equivalent of exponents
y*log(6) = log(7)
:
.778y = .845
y = .845%2F.778
y = 1.086 ~ 1.09
:
You can check this on a good calc: Enter: 6^1.086, = 6.999 ~ 7
n
:
Did this help you?

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Here's a way that it can be done.
.
Given:
.
y+=log%286%2C3.5%29+%2B+log%286%2C2%29
.
By the rules of logarithms, if you have the sum of two logs (must be the same base) you
can apply the product rule:
.
y+=+log%28b%2CA%29+%2B+log%28b%2CC%29+=+log%28b%2C%28A%2AC%29%29
.
By comparing this rule term by term with the given problem, you can see that b = 6, A = 3.5, and C = 2
.
Substituting these values into the rule, the problem rule gives:
.
y+=+log%286%2C3.5%29+%2B+log%286%2C2%29+=+log%286%2C%283.5%2A2%29%29
.
and when you multiply out the two terms you get:
.
y+=+log%286%2C7%29
.
Next you can convert this logarithmic for to exponential form by applying the conversion rule:
.
y+=+log%28b%2CD%29 is equivalent to the exponential form b%5Ey+=+D%29
.
If you compare this form to the logarithmic form you now have you can see that y = y, b = 6, and D = 7
.
Substituting these values into the exponential form you convert your logarithmic form into the
exponential form:
.
6%5Ey+=+7
.
Now take the log to the base 10 of both sides of this exponential form and you get:
.
log%2810%2C6%5Ey%29+=+log%2810%2C7%29
.
Applying the exponential rule to the left side brings the exponent out as a multiplier and
the equation then becomes:
.
y%2Alog%2810%2C6%29+=+log%2810%2C7%29
.
You can now use a calculator to find log%2810%2C6%29. Enter 6 and press the "log" key. You should
find that the value you get for this log is 0.77815125.
.
Similarly you can find log%2810%2C7%29 by entering 7 and pressing the "log" key to get that
this log is 0.84509804.
.
Substitute the two values you have for the base 10 logs of 6 and 7 into the equation and
the equation becomes:
.
y%2A%280.77815125%29+=+0.84509804
.
Solve for y by dividing both sides by 0.77815125 and you get:
.
y+=+0.84509804%2F0.77815125+=+1.086033133
.
Since way back at the beginning of the problem we defined y as log%286%2C3.5%29+%2B+log%286%2C2%29
you can now say that:
log%286%2C3.5%29+%2B+log%286%2C2%29+=+1.086033133
.
and this rounds off to:
.
log%286%2C3.5%29+%2B+log%286%2C2%29+=+1.09
.
The answer is answer A ... 1.09
.
Hope this helps you to find where you went astray in working the problem.
.