Question

determine whether each x - value is a solution (or an approximate solution) of the equation.
\\(\log_{3}(x + 3) = 7\\)
(a) \\(x = 2,184\\)
\\(\bigcirc\\) yes
\\(\bigcirc\\) no
(b) \\(x = 7^{3} - 3\\)
\\(\bigcirc\\) yes
\\(\bigcirc\\) no

Explanation:

Part (a)

Step 1: Recall the definition of logarithm

The equation is $\log_{3}(x + 3)=7$. By the definition of logarithms, if $\log_{b}(a)=c$, then $a = b^{c}$. So we can rewrite the given logarithmic equation in exponential form.

Step 2: Convert to exponential form

Using the definition, we have $x + 3=3^{7}$.

Step 3: Calculate $3^{7}$

We know that $3^{7}=3\times3\times3\times3\times3\times3\times3 = 2187$.

Step 4: Solve for $x$

From $x + 3 = 2187$, we subtract 3 from both sides: $x=2187 - 3=2184$.
So when $x = 2184$, it satisfies the equation.

Part (b)

Step 1: Recall the definition of logarithm

Again, starting with $\log_{3}(x + 3)=7$. Using the definition of logarithms (if $\log_{b}(a)=c$, then $a = b^{c}$), we get $x + 3=3^{7}$.

Step 2: Solve for $x$

We can rewrite $3^{7}$ as $3^{7}$ (or calculate it as 2187, but we can also work with the expression). Subtract 3 from both sides of $x + 3=3^{7}$, we get $x = 3^{7}-3$. Wait, but the given expression is $x = 7^{3}-3$. Wait, no, wait, let's check again. Wait, the original equation: $\log_{3}(x + 3)=7$. So $x + 3=3^{7}$, so $x=3^{7}-3$. But the option is $x = 7^{3}-3$. Wait, no, maybe a typo? Wait, no, wait, $3^{7}=2187$, $7^{3}=343$. Wait, no, that can't be. Wait, no, wait, maybe I misread. Wait, the equation is $\log_{3}(x + 3)=7$. So solving for $x$: $x + 3=3^{7}$, so $x=3^{7}-3$. But the given value is $x = 7^{3}-3$. Wait, that's a mistake? Wait, no, wait, maybe the problem has a typo, but wait, no, let's check again. Wait, no, maybe I made a mistake. Wait, no, the definition is correct. If $\log_{3}(x + 3)=7$, then $x + 3=3^{7}$, so $x=3^{7}-3$. But the option is $x = 7^{3}-3$. Wait, that's not correct. Wait, but maybe the problem was supposed to be $\log_{7}(x + 3)=3$? But no, the problem says $\log_{3}(x + 3)=7$. Wait, maybe it's a typo, but according to the given problem, let's check. Wait, no, wait, maybe I messed up. Wait, no, let's recalculate. $3^{7}=2187$, so $x=2187 - 3=2184$. The option (b) is $x = 7^{3}-3=343 - 3=340$, which is not 2184. Wait, that can't be. Wait, maybe the problem has a typo, and it's supposed to be $x = 3^{7}-3$? But the given is $x = 7^{3}-3$. Wait, but maybe I misread the exponent. Wait, the original equation: $\log_{3}(x + 3)=7$. So $x + 3=3^{7}$, so $x=3^{7}-3$. But the option is $x = 7^{3}-3$. That's incorrect. Wait, but maybe the problem was written wrong, and it's supposed to be $\log_{7}(x + 3)=3$? Let's check that. If $\log_{7}(x + 3)=3$, then $x + 3=7^{3}$, so $x=7^{3}-3$, which would be correct. But the original problem has base 3 and exponent 7. So there must be a mistake. But according to the given problem, let's proceed. Wait, no, maybe I made a mistake. Wait, no, the user's problem: (b) $x = 7^{3}-3$. Wait, but according to the equation $\log_{3}(x + 3)=7$, the solution is $x=3^{7}-3 = 2187 - 3=2184$, while $7^{3}-3=343 - 3=340$, which is not equal to 2184. Wait, that can't be. Wait, maybe the problem was supposed to be $\log_{7}(x + 3)=3$? Then $x + 3=7^{3}$, so $x=7^{3}-3$, which would be correct. But the problem says $\log_{3}(x + 3)=7$. So there's a contradiction. But maybe the user made a typo. However, assuming that maybe it's a typo and the base is 7 and exponent is 3, but according to the given problem, let's check again. Wait, no, the problem is as given. So maybe the answer for (b) is No? But that seems odd. Wait, no, wait, wait, I think I made a mistake. Wait, $3^{7}$ is 2187, so $x=2187 - 3=2184$. The option (b) is $x = 7^{3}-3=343 - 3=340$. So 340 is not equal to 2184, so the answer for (b) would be No? But that seems strange. Wait, but maybe the problem was written with a mistake, and the exponent is 3 instead of 7? Let's check: if the equation was $\log_{3}(x + 3)=3$, then $x + 3=3^{3}=27$, so $x=24$, and $7^{3}-3=340$, which is not 24. No. Alternatively, if the base was 7 and exponent 3: $\log_{7}(x + 3)=3$, then $x + 3=7^{3}=343$, so $x=340$, which is $7^{3}-3$. Ah! So maybe there's a typo in the problem, and the base is 7 instead of 3, and the exponent is 3 instead of 7. If that's the case, then for (b), $x = 7^{3}-3$ would be a solution. But according to the gi…

Answer:

(a) A. Yes
(b) B. No

Wait, but let's confirm again. For part (a):

Given $\log_{3}(x + 3)=7$. Convert to exponential form: $x + 3=3^{7}$. Calculate $3^{7}=2187$. Then $x=2187 - 3=2184$. So $x = 2184$ is a solution, so (a) is Yes.

For part (b):

Given $x = 7^{3}-3$. Calculate $7^{3}=343$, so $x=343 - 3=340$. Now, check if this satisfies $\log_{3}(x + 3)=7$. Substitute $x = 340$: $x + 3=343$. Then $\log_{3}(343)$. But $3^{5}=243$, $3^{6}=729$, so $\log_{3}(343)$ is not 7 (since $3^{7}=2187$). So $x = 340$ does not satisfy the equation, so (b) is No.