# ln(35 ```M305G - Worksheet - 10/5
1. Let ln(5) = c and ln(7) = d. Write the following expressions in terms of c and d.
1
a) ln( 35
) = ln(35−1 ) = −ln(35) = −ln(5 &middot; 7) = −(ln(5) + ln(7)) = −(c + d)
b) ln(625) = ln(54 ) = 4ln(5) = 4c
q
1
c) ln( 3 57 ) = ln(( 57 ) 3 ) = 13 ln( 75 ) = 13 (ln(5) − ln(7)) = 31 (c − d)
2. Use the change of base formula to find the exact value of
log5 (16)
log5 (2)
log2 (2)
2 (16)
= ( log
)/( log
)=
log2 (5)
2 (5)
log2 (16)
log2 (2)
log5 (16)
.
log5 (2)
= log2 (16) = log2 (24 ) = 4
3. Find the exact value of log2 (5)&middot;log5 (4). Can you write down a general fact about products
like these?
log2 (4)
log2 (5) &middot; log5 (4) = log2 (5) &middot; ( log
) = log2 (4) = 2
2 (5)
In general, loga (b) &middot; logb (c) = loga (c). This IS the base change formula.
3
√
x+1
4. Let f (x) = log[ x(x−2)
2 ]. Use properties of logs to rewrite f as the sum or difference of
logs. This means that, in the answer, the arguments of the logs are as simple as possible.
What is the domain of f ? What is the domain of the result of your expansion.
√
√
3 x+1
3
x + 1) − log[(x − 2)2 ] = 3log(x) + 12 log(x + 1) −
f (x) = log[ x(x−2)
2 ] = log(x ) + log(
2log(x − 2)
3
√
x+1
The domain of f is x such that x(x−2)
2 &gt; 0. First, for the square root to makes sense,
x &gt; −1, and x 6= 2 for the denominator. But then the square root and the denominator
ar always positive, so we’ve really only got to solve x3 &gt; 0, which is x &gt; 0. Hence Df is
(0, 2) ∪ (2, ∞).
But if we call g(x) = 3log(x) + 21 log(x + 1) − 2log(x − 2) the resulting function, we need
x &gt; 0, x &gt; −1 and x &gt; 2. Combining these gives x &gt; 2, otherwise written Dg = (2, ∞).
So we lost one of the chunks of our domain.
5. Write the following expressions as a single logarithm.
1
3
1
3
(x +1)
1
1
+ 1) − 2ln(2x − 1) − ln(x) = ln[(x3 + 1) 3 ] + ln[ (2x−1)
2 ] + ln( x ) = ln[ x(2x−1)2 ]
√
1
2
b) 21log( 3 x)+log(9x2 )−log(3) = log[(x 3 )21 ]+log( 9x3 ) = log(x7 )+log(3x2 ) = log(3x9 )
a)
1
ln(x3
3
6. Solve the following equations for x.
a)
1
log234576 (x)
2
= 2log234576 (2)
√
→ log234576 ( x) = log234576 (22 )
√
↔ x = 4, so x = 16, which makes sense in the original equation
b) 2log3 (x + 4) − log3 (9) = 2
→ log3 [(x + 4)2 ] − 2 = 2
→ log3 [(x + 4)2 ] = 4
↔ (x + 4)2 = 34
→ x + 4 = &plusmn;9
→ x = 5, −13, but only x = 5 is in the domain.
c) 22x + 2x+2 − 12 = 0
→ (2x )2 + 22 (2x ) − 12 = 0, so let u = 2x
→ u2 + 4u − 12 = (u + 6)(u − 2) = 0, so u = 2, −6
→ 2x = 2, which gives x = 1 or 2x = −6, which gives no solution. x = 1 makes
sense in the original equation, so it is our only solution.
d) ( 34 )1−x = 5x
→ ln[( 43 )1−x ] = ln(5x )
→ (1 − x)ln( 34 ) = xln(5)
→ ln( 43 ) − xln( 34 ) = xln(5)
)
→ ln( 43 ) = xln(5) + xln( 34 ) = x(ln(5) + ln( 34 )) = xln( 20
3
→
ln( 34 )
ln( 20
)
3
= x, which makes sense in the original equation.
e) loga (x − 1) − loga (x + 6) = loga (x − 2) − loga (x + 3)
) = loga ( x−2
)
→ loga ( x−1
x+6
x+3
x−2
= x+3
→ x−1
x+6
→ (x − 1)(x + 3) = x2 + 2x − 3 = x2 + 4x − 12 = (x − 2)(x + 6)
→ 0 = 2x − 9, so x = 92 , which makes sense in the original equation.
f) log2 (x + 1) − log4 (x) = 1 No, the 4 is not a typo.
2 (x)
=1
→ log2 (x + 1) − log
log2 (4)
→ log2 (x + 1) − log22(x) = 1
√
→ log2 (x + 1) − log2 ( x) = 1
√ ) = 1
→ log2 ( x+1
x
→
x+1
√
x
= 21
√
→x+1=2 x
→ (x + 1)2 = x2 + 2x + 1 = 4x
→ x2 − 2x + 1 = (x − 1)2 = 0, so x = 1, which makes sense in the original equation.
g) 3x + 3−x = 2
→ 3x − 2 + 3−x = 0
→ (3x )2 − 2(3x ) + 3−x 3x = (3x )2 − 2(3x ) + 1 = 0, so let u = 3x
→ u2 − 2u + 1 = (u − 1)2 = 0, so u = 1
→ 3x = 1, so x = 0, which makes sense in the original equation.
h) log2 [xlog2 (x) ] = 4
→ (log2 (x))(log2 (x))(log2 (x))2 = 4
→ log2 (x) = &plusmn;2
so either log2 (x) = 2 ↔ x = 22 = 4
or log2 (x) = −2 ↔ x = 2−2 = 14 , both of which make sense in the original equation.
7. Express y as a functions of x, where C is just a constant.
a) ln(y) = ln(x) + ln(C)
→ ln(y) = ln(xC) ↔ y = xC
b) ln(y) = ln(x + C)
↔y =x+C
c) ln(y + 4) = 5x + ln(C)
→ eln(y+4) = e5x+ln(C) = e5x eln(C)
→ y + 4 = e5x C
→ y = Ce5x − 4
Notice that I’ve been writing “which makes sense in the original equation”, but I’m not
showing you why I think that. You’ll need to do that, so here are three examples:
For 6h, log2 [xlog2 (x) ] = 4, we got x = 4, 14 . Let’s start with 4. We can plug 4 into that
exponent, and it will spit out 2. The we can raise 4 to the power 2, and the result is in the
domain of log2 . Similarly, if we plug 14 into the exponent, we get some number. We can raise
1
to that number to get something positive, so it is also in the domain of log2 .
4
Or, for 6e, loga (x − 1) − loga (x + 6) = loga (x − 2) − loga (x + 3), we got x = 29 . We check that
each logarithm in the original equation makes sense. For that , we must be putting in positive
things. Is 92 − 1 positive? Yes. Is 92 + 6 positive? Yes. Is 92 − 2 positive? Yes. Is 29 + 3 positive?
Yes. So this number makes sense in the original equation.
Finally, for 3g, 3x +3−x = 2, we got x = 1. Any number makes sense in the original equation
because the domain of all exponentials is R, so x = 1 definitely makes sense.
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