Expert: Josh - 6/7/2006

Question
I need to solve 3 different log problems.



1)                 log(sub)8(n-3)+log(sub)8(n+4)=1



                

2)                     log(sub)4(3x-1)=(4x+3)





3)                     log(sub)2(32)=3x





Also I have to rewrite this one problem as a single log function:



1)                      1/5(log63-log7)+2log5





Last I have to graph and then find the domain, range,and y-intercepts:



1)                 y=log(sub)10(x)





2)                 Log(sub)10(x+2)



I dont know how to even go about these problems.

Answer
Okay Alex, let's go through these one-by-one.

PART ONE
In Q1, we have to find the unknown "n" which satisfies the equation log_8(n-3)+log_8(n+4)=1.

The key results that you should remember are:
log(a)+log(b)=log(ab)
log(a)-log(c)=log(a/c)

Using the first logarithmic rule,
log_8(n-3)+log_8(n+4)=1
log_8((n-3)*(n+4))=1, [read note 1]
(n-3)*(n+4)=8 ...now you get a quadratic equation
n^2+n-12=8
n^2+n-20=0 ...which we factorize as
(n-4)(n+5)=0
so, this problem has two possible solutions: n=4, or n=-5. We rule out n=-5, since if n=-5, log(n+4) (with negative argument) would be undefined.

note 1: Roughly speaking, to "reverse" the effects of the logarithmic function, on both sides of the equation, we need to raise eight to the power of the corresponding (left/right) expression.

Here is a simple illustration. Imagine we have a quantity, y=log_2(x). It doesn't matter what "x" is, if we raise 2 to the power of y, 2^(log_2(x))=x. We say that 2^y inverts the logarithmic function y=log_2(x); in the sense that it returns the original input x. So, if you get the drift, 8^y inverts the function y=log_8(x); in the sense that 8^(log_8(x))=x.

Q2) If you copied the question right, this problem has no solution.

(i) The log(3x-1) part is defined only if the argument is strictly positive. Thus, we look for a solution where x > 1/3.

(ii) For x>1/3, the logarithmic function does not intersect with y=4x+3.

3) This one is much easier. 32 is divisible by 2. Since 2^5=32, log_2(32)=5. You solve for 5=3x.

==========
Q1) To write 1/5(log63-log7)+2log5 as a single log function, apply the rules given earlier in PART ONE Q1.
(1/5)*(log(63)-log(7))+2*log(5)
=(1/5)*(log(63/7))+2*log(5)...using log(a)-log(b)=log(a/b)
=(1/5)*(log(63/7))+log(5^2)...using clog(a)=log(a^c)
=(1/5)*(log(9))+log(25)
=log(9^(1/5))+log(25) ...using (1/k)log(h)=log(h^(1/k))
=log(9^(1/5)*25)...using log(d)+log(e)=log(de)

1) y=log(sub)10(x)
I can't draw graphs here, you'll have to look at your textbook to figure what a typical log function looks like.

The "domain" consists of all legitimate x values for which the function is defined. Since we cannot raise 10 to any power to obtain a negative number "x", x must be positive. Ans: x>0.

Range: y-values that can be attained by the function. For y=log_10(x), y is unbounded. It can take value from -infinity to +infinity. As x gets closer to the vertical asymptote x=0 (approach this from the right hand side), the log function goes toward -infinity (some infinitely large negative value). Although the rate of change slows down significantly as x increases (remember that the derivative is 1/x), it can in theory approach +infinity (some arbitrary, infinitely large positive value) if x is allowed to grow without bound.

y-intercepts: where the curve crosses the y-axis; this is usually found by setting x=0. Here, we are forbidden to do this, as the function is not defined at x=0. x=0 represents a limit. So, there is no y-intercept here.

2) With Log_10(x+2), it has the same shape as log_10(x), except it is translated horizontally (shifted to the left) by 2 units along the x-axis.

Domain: In this case, we need (x+2)>0, which implies x>-2.

Range: Same as previous question
Y-intercept: set x=0 in y=log_10(x+2)
X-intercept: solve for x when y=log_10(x+2)=0
This gives x+2=10^0=1. Function crosses at x=-1.