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leafsequallove · Jun 11, 2009, 12:23 PM

I have my math provincial exam tomorrow, and I forget how to do logarithms.

for example how would you go about solving the problem
log_2(3x)+log_2(10)=log_3(81)

Perito · Jun 11, 2009, 12:37 PM

Here's a link for reviewing some logarithm definitions:

Mathwords: Logarithm Rules

Basically, to solve these problems, you isolate the variable and if it's in the exponent, you take the logarithm of both sides of the equation.

$a^x=b$

$log(a^x)=b$

$x\,log(a)=b$

$x=\frac {b}{log(a)}$

If you have a term involving the log of something, you may have to take an antilog -- as in your example:

---------------------------

$log_2(3x)+log_2(10)=log_3(81)$

The right side of the equation is simple if you recognize that 81 is 9x9 = 3x3x3x3

$log_3(81) = log_3(3^4) = 4log_3(3)=4$

The second term isn't quite as easy.

$log_2(10)=x$ , but this can be rewritten (using the definition of logarithms -- this is the same as the antilog)

$10 = 2^x$

$log(2^x)=log(10)$ This is the common log -- most calculators can calculate common logs (base 10) or natural logs (base e), but they can't ordinarily calculate logs to some random base.

$xlog(2) = 1$

$x=\frac {1}{log(2)} = A \approx 3.321928$

back to the original problem. I'm going to use the A that I wrote above to represent the irrational number.

$log_2(3x)+log_2(10)=log_3(81)$

$log_2(3x)+A=4$

$log_2(3x)=(4-A)$

take the antilog (raise 2 to the power of each side, since this is a base 2 logarithm)

$2^{log_2(3x)}=3x=2^{(4-A)}$

and therefore $x = \frac {2^{(4-A)}}{3}$

Remember that A is simply the irrational constant $1 / log(2)$

jorgepmichel · Oct 25, 2009, 11:06 AM

how do you get A=3.321928 from x= 1/log(2)?

Chris-infj · Oct 25, 2009, 11:29 AM

$log_2(3x) + log_2(10) = log_3(81)$

Another way of doing it is to simplify the entire left hand side to this

$log_2(3x(10))$ using the rule that $log_a(x) + log_a(y) = log_a(xy)$ that is the sum of the logarithms to the same base is equal to the logarithm of the product to the same base.

The right hand side of the equation is 4 as posted previously.