ln52ln4+log(x2+8)51=log5(7x2+9x)
x=?
\( \frac{2\ln4}{\ln5}+\frac{1}{\log_{(x^2+8)}5}=\log_5(7x^2+9x) \)
\( x=\text{?} \)
\( x\ln7= \)
\( \frac{\log_{4x}9}{\log_{4x}a}= \)
\( \frac{\log_89a}{\log_83a}= \)
\( (\log_7x)^{-1}= \)
To solve the given equation, follow these steps:
We start with the expression:
Use the change-of-base formula to rewrite everything in terms of natural logarithms:
Multiplying the entire equation by to eliminate the denominators:
By properties of logarithms (namely the product and power laws), combine the left side using the addition property:
Since the natural logarithm function is one-to-one, equate the arguments:
Rearrange this into a standard form of a quadratic equation:
Attempt to solve this quadratic equation using the quadratic formula:
Where , , and .
Calculate the discriminant:
The discriminant is positive, suggesting real solutions should exist, however, verification against the domain constraints of logarithms (arguments must be positive) is needed.
After solving , the following is noted:
The polynomial does not yield any values in domains valid for the original logarithmic arguments.
Cross-verify the potential solutions against original conditions:
Solutions obtained do not satisfy these together within the purview of the rational roots and ultimately render no real value for .
Therefore, the solution to the problem is: There is no solution.
No solution
\( n\log_xa= \)
\( x\log_m\frac{1}{3^x}= \)
\( \ln4x= \)
\( \frac{\frac{2x}{\log_89}}{\log_98}= \)
\( \frac{4a^2}{\log_79}\colon\log_97=16 \)
Calculate a.
Calculate a.
Solve for X:
\( \log_3(x+2)\cdot\log_29=4 \)
Calculate X:
\( 2\log(x+4)=1 \)
\( 2\log(x+1)=\log(2x^2+8x) \)
\( x=\text{?} \)
\( \frac{1}{2}\log_3(x^4)=\log_3(3x^2+5x+1) \)
\( x=\text{?} \)
\( \frac{\log_4(x^2+8x+1)}{\log_48}=2 \)
\( x=\text{?} \)
Solve for X:
Calculate X:
Find X
\( \frac{\log_84x+\log_8(x+2)}{\log_83}=3 \)
\( \frac{2\log_7(x+1)}{\log_7e}=\ln(3x^2+1) \)
\( x=\text{?} \)
\( x=\text{?} \)
\( \log_{\frac{1}{2}}5-\log_{\frac{1}{2}}4\le\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}3 \)
\( \log_35x\times\log_{\frac{1}{7}}9\ge\log_{\frac{1}{7}}4 \)
\( \log_{\frac{1}{3}}e^2\ln x<3\log_{\frac{1}{3}}2 \)
Find X
0 < x\le3.75
0 < x\le\frac{1}{245}
\log_{\frac{1}{3}}e^2\ln x<3\log_{\frac{1}{3}}2
\sqrt{8} < x