Question

Logarithms- How to answer these questions?

Answer 1

For #3:
`\log_px=-4`

Explanation:

I'm a little rusty on my logarithm rules for #2, but here's an explanation of #3.

Logarithms: the Inverse of Exponents

In a sense, we can think of operations like subtraction and division as different ways of representing addition and multiplication. For instance, the same relationship described by the equation 2 + 3 = 5 is captured in the equation 5 - 3 = 2, and 5 × 2 = 10 can be restated as 10 ÷ 2 = 5 without any loss of meaning.

Logarithms do the same thing for exponents: the expression
`2^3=8` can be expressed in logarithms as
`\log_28=3`. Put another way, logarithms are a sort of way of pulling an exponent out onto its own side of the equals sign.

The Problem

Our problem gives us two facts to start: that
`log_p(x^2y^3)=7` and
`p^5=y`. With that, we're expected to find the value of
`\log_px`.
`p^5=y` stands out as the odd-equation-out here; it's the only one not in terms of logarithms. We can fix that by rewriting it as the equivalent statement
`log_py=5`. Now, let's unpack that first logarithm.

Justifying Some Logarithm Rules

For a refresher, let's talk about some of the rules logarithms follow and why they follow them:

Product Rule:
`\log_b(MN)=\log_bM+\log_bN`

The product rule turns multiplication in the argument (parentheses) of a logarithm into addition. For a proof of this, consider two numbers
`M=b^x` and
`N=b^y`. We could rewrite these two equations with logarithms as
`\log_bM=x` and
`\log_bN=y`. With those in mind, we could say the following:

• 
  `\log_b(MN)=\log_b(b^xb^y)` (Substitution)
• 
  `\log_b(b^xb^y)=log_b(b^(x+y))` (Laws of exponents)
• 
  `\log_b(b^(x+y))=x+y` (
  `\log_b(b^n)=n`)
• 
  `x+y=\log_bM+\log_bN` (Substitution)

And we have our proof.

Exponent Rule:
`\log_b(M^n)=n\log_bM`

Since exponents can be thought of as abbreviations for repeated multiplication, we can rewrite
`\log_b(M^n)` as
`\log_b(M* M\cdots * M)`, where M is being multiplied by itself n times. From there, we can use the product rule to rewrite our logarithm as the sum
`\log_bM+\log_bM+\cdots+\log_bM`, and since we have the term
`\log_bM` added n times, we can rewrite is as
`n\log_bM`, proving the rule.

Solving the Problem

With those rules in hand, we're ready to solve the problem. Looking at the equation
`\log_p(x^2y^3)=7`, we can use the product rule to split the logarithm into the sum
`\log_p(x^2)+\log_p(y^3)=7`, and then use the product rule to turn the exponents in each logarithm's argument into coefficients, giving the equation
`2\log_px+3\log_py=7`.

Remember how earlier we rewrote
`p^5=y` as
`log_py=5`? We can now use that fact to substitute 5 in for
`log_py`, giving us

`2\log_px+3(5)=7`

From here, we can simply solve the equation for
`\log_px`:

`2\log_px+15=7\\2\log_px=-8\\\\\log_px=-4`

Answer 2

2. m = b³ (= 216)

3. logp(x) = -4

Explanation:

2. The given equation can be written using the change of base formula as ...

... log(m)/log(b) + 9·log(b)/log(m) = 6

If we define x = log(m)/log(b), then this becomes ...

... x + 9/x = 6

Subtracting 6 and multiplying by x gives ...

... x² -6x +9 = 0

... (x -3)² = 0 . . . . . factored

... x = 3 . . . . . . . . . value of x that makes it true

Remembering that x = log(m)/log(b), this means

... 3 = log(m)/log(b)

... 3·log(b) = log(m) . . . . . multiply by the denominator; next, take the antilog

... m = b³ . . . . . . the expression you're looking for

___

3. Substituting the given expression for y, the equation becomes ...

... logp(x^2·(p^5)^3) = 7

... logp(x^2) + logp(p^15) = 7 . . . . . use the rule for log of a product

... 2logp(x) + 15 = 7 . . . . . . . . . . . . . use the definition of a logarithm

... 2logp(x) = -8 . . . . . . . . . . . . . . . . subtract 15

... logp(x) = -4 . . . . . . divide by 2