Respuesta :

sqdancefan

Answer:

2. m = b³ (= 216)

3. logp(x) = -4

Step-by-step 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

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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

highwireact

Answer:

For #3: [tex]\log_px=-4[/tex]

Step-by-step 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 [tex]2^3=8[/tex] can be expressed in logarithms as [tex]\log_28=3[/tex]. 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 [tex]log_p(x^2y^3)=7[/tex] and [tex]p^5=y[/tex]. With that, we're expected to find the value of [tex]\log_px[/tex]. [tex]p^5=y[/tex] 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 [tex]log_py=5[/tex]. 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: [tex]\log_b(MN)=\log_bM+\log_bN[/tex]

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

  • [tex]\log_b(MN)=\log_b(b^xb^y)[/tex] (Substitution)
  • [tex]\log_b(b^xb^y)=log_b(b^{x+y})[/tex] (Laws of exponents)
  • [tex]\log_b(b^{x+y})=x+y[/tex] ([tex]\log_b(b^n)=n[/tex])
  • [tex]x+y=\log_bM+\log_bN[/tex] (Substitution)

And we have our proof.

Exponent Rule: [tex]\log_b(M^n)=n\log_bM[/tex]

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

Solving the Problem

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

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

[tex]2\log_px+3(5)=7[/tex]

From here, we can simply solve the equation for [tex]\log_px[/tex]:

[tex]2\log_px+15=7\\2\log_px=-8\\\\\log_px=-4[/tex]

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