MHB What is the value of x in the equation 7^(x) + 3^(x) = (2.1)^(-1)?

[mathdad]

Solve for x:

7^(x) + 3^(x) = (2.1)^(-1)

My Work:

Note: I understand (2.1)^(-1) as the decimal number 2.1 raised to the negative 1, which can be expressed as 1/(2.1).

7^(x) + 3^(x) = (2.1)^(-1)

ln[7^x] + ln[3^x] = ln[1/(2.1)]

xln7 + xln3 = -0.741937

x(ln7 + ln3) = -0.741937

x = 1/[(-0.741937)(ln7 + ln3)]

Use a calculator to find an approximate x value.

x = -0.442704

Correct?

 

[Opalg]

Solve for x:

7^(x) + 3^(x) = (2.1)^(-1)

My Work:

Note: I understand (2.1)^(-1) as the decimal number 2.1 raised to the negative 1, which can be expressed as 1/(2.1).

7^(x) + 3^(x) = (2.1)^(-1)

ln[7^x] + ln[3^x] = ln[1/(2.1)]

xln7 + xln3 = -0.741937

x(ln7 + ln3) = -0.741937

x = 1/[(-0.741937)(ln7 + ln3)]

Use a calculator to find an approximate x value.

x = -0.442704

Correct?

Not correct. Look at those two lines in red. What is wrong there?

A better approach would be to start from the fact that $$2.1^{-1} = \frac1{2.1} = \frac{10}{21}.$$ Notice that $10 = 3+7$ and $21 = 3\times7$.

 

[mathdad]

Not correct. Look at those two lines in red. What is wrong there?

A better approach would be to start from the fact that $$2.1^{-1} = \frac1{2.1} = \frac{10}{21}.$$ Notice that $10 = 3+7$ and $21 = 3\times7$.

I am wrong. I get it. Can you provide the needed steps for me to solve this exponential equation?

 

[MarkFL]

I am wrong. I get it. Can you provide the needed steps for me to solve this exponential equation?

You see the LHS of the equation is the sum of 7 and 3 raised to the same power. Using the suggestion posted by Opalg, can you express 10/21 as the sum of 7 and 3 raised to the same power?

 

[mathdad]

You see the LHS of the equation is the sum of 7 and 3 raised to the same power. Using the suggestion posted by Opalg, can you express 10/21 as the sum of 7 and 3 raised to the same power?

I do not understand.

 

[MarkFL]

I do not understand.

Okay, we are originally given:

$$7^x+3^x=2.1^{-1}$$

We can write this as:

$$7^x+3^x=\frac{10}{21}=\frac{3+7}{7\cdot3}=\frac{1}{7}+\frac{1}{3}=7^{-1}+3^{-1}$$

And so what value must $x$ have?

 

[mathdad]

Okay, we are originally given:

$$7^x+3^x=2.1^{-1}$$

We can write this as:

$$7^x+3^x=\frac{10}{21}=\frac{3+7}{7\cdot3}=\frac{1}{7}+\frac{1}{3}=7^{-1}+3^{-1}$$

And so what value must $x$ have?

The value of x = (1/7) + (1/3), right?

 

[MarkFL]

The value of x = (1/7) + (1/3), right?

No, we have:

$$7^x+3^x=7^{-1}+3^{-1}$$

Therefore, the exponents must be equal, which means:

$$x=-1$$

 

[mathdad]

Thanks.