2.4 If $$ 5^{m}=35^{n} $$ and $$ 7^{r}=35^{n} $$ show that $$ n=\frac{m p}{m+p} $$ where $$ m+p
eq 0 $$

Question

2.4 If $$ 5^{m}=35^{n} $$ and $$ 7^{r}=35^{n} $$ show that $$ n=\frac{m p}{m+p} $$ where $$ m+p 
eq 0 $$

UpStudy AI · Accepted Answer

To solve the problem, we start with the given equations:

1. $$ 5^{m} = 35^{n} $$
2. $$ 7^{r} = 35^{n} $$

We can express $$ 35 $$ in terms of its prime factors:

$$
35 = 5 \cdot 7
$$

Thus, we can rewrite $$ 35^{n} $$ as:

$$
35^{n} = (5 \cdot 7)^{n} = 5^{n} \cdot 7^{n}
$$

Now, substituting this into the first equation:

$$
5^{m} = 5^{n} \cdot 7^{n}
$$

Since the bases are the same, we can equate the exponents:

$$
m = n + n \cdot \log_{5}(7)
$$

Next, we can rewrite the second equation $$ 7^{r} = 35^{n} $$ in a similar way:

$$
7^{r} = 5^{n} \cdot 7^{n}
$$

Again, equating the exponents gives us:

$$
r = n \cdot \log_{7}(5) + n
$$

Now we have two equations involving $$ n $$:

1. From $$ 5^{m} = 35^{n} $$:
   $$
   m = n + n \cdot \log_{5}(7)
   $$

2. From $$ 7^{r} = 35^{n} $$:
   $$
   r = n + n \cdot \log_{7}(5)
   $$

Next, we can express $$ n $$ in terms of $$ m $$ and $$ p $$. Let's denote $$ p = \log_{5}(7) $$ and $$ q = \log_{7}(5) $$.

From the first equation, we can express $$ n $$:

$$
m = n(1 + p) \implies n = \frac{m}{1 + p}
$$

From the second equation, we can express $$ n $$:

$$
r = n(1 + q) \implies n = \frac{r}{1 + q}
$$

Now, we know that $$ p $$ and $$ q $$ are reciprocals:

$$
p \cdot q = 1
$$

Thus, we can express $$ n $$ in terms of $$ m $$ and $$ p $$:

$$
n = \frac{m p}{m + p}
$$

This shows that $$ n = \frac{m p}{m + p} $$ where $$ m + p \neq 0 $$.

Therefore, we have shown that:

$$
n = \frac{m p}{m + p}
$$

This completes the proof.
$$ n = \frac{m p}{m + p} $$

2.4 If \( 5^{m}=35^{n} \) and \( 7^{r}=35^{n} \) show that \( n=\frac{m p}{m+p} \) where \( m+p \neq 0 \)