Question

use the elementary properties of logarithms to solve the following equation. write your answer as a fraction reduced to lowest terms.

$7^{\log_{7}(x)} = 49$

answer how to enter your answer (opens in new window)
4 points

$x = $

Explanation:

Step1: Rewrite 49 as a power of 7

Since \( 49 = 7^2 \), the equation \( 7^{\log_{7}(x)} = 49 \) can be rewritten as \( 7^{\log_{7}(x)} = 7^2 \).

Step2: Use the property of exponents

If \( a^m = a^n \), then \( m = n \) (for \( a>0, a
eq1 \)). Here, \( a = 7 \), \( m=\log_{7}(x) \), and \( n = 2 \). So we have \( \log_{7}(x)=2 \).

Step3: Convert logarithm to exponential form

The logarithmic equation \( \log_{a}(b)=c \) is equivalent to \( a^c = b \). For \( \log_{7}(x)=2 \), we get \( x = 7^2 \).

Step4: Calculate \( 7^2 \)

\( 7^2=49 \)? Wait, no, wait. Wait, no, let's check again. Wait, no, the property of logarithms: \( a^{\log_{a}(x)}=x \). Oh! I made a mistake in Step2. Let's correct that.

Wait, the correct property: \( a^{\log_{a}(x)}=x \). So the left - hand side of the equation \( 7^{\log_{7}(x)} \) is just \( x \). So the equation \( 7^{\log_{7}(x)} = 49 \) simplifies directly to \( x = 49 \)? Wait, no, that can't be. Wait, no, wait the original equation: \( 7^{\log_{7}(x)}=49 \). By the inverse property of logarithms and exponents, \( a^{\log_{a}(y)}=y \). So here, \( a = 7 \) and \( y=x \), so \( 7^{\log_{7}(x)}=x \). Therefore, the equation becomes \( x = 49 \)? Wait, but let's verify. If \( x = 49 \), then \( \log_{7}(49)=\log_{7}(7^2) = 2 \), and \( 7^{2}=49 \), so \( 7^{\log_{7}(49)}=7^{2}=49 \), which matches the right - hand side. Wait, but maybe I misread the logarithm base. Wait, the logarithm is \( \log_{7}(x) \), right? The original equation is \( 7^{\log_{7}(x)} = 49 \). So by the identity \( a^{\log_{a}(x)}=x \), we have \( x = 49 \)? Wait, no, that seems too simple, but let's check the exponent property again.

Yes, the property \( a^{\log_{a}(x)}=x \) holds for \( x>0 \) and \( a>0,a
eq1 \). So in this case, \( a = 7 \), so \( 7^{\log_{7}(x)}=x \), so \( x = 49 \). Wait, but let's do it step by step again.

Alternative approach:

Let \( y=\log_{7}(x) \), then the equation is \( 7^{y}=49 \). Since \( 49 = 7^{2} \), we have \( 7^{y}=7^{2} \), so \( y = 2 \). But \( y=\log_{7}(x) \), so \( \log_{7}(x)=2 \). Converting to exponential form, \( x = 7^{2}=49 \). Yes, that's correct.

Wait, but earlier when I thought of the identity \( a^{\log_{a}(x)}=x \), that's a direct way. So \( 7^{\log_{7}(x)}=x \), so \( x = 49 \).

Answer:

\( x=\boxed{49} \)