log7x+log(x+1)−log7=log2x−logx
?=x
\( \log7x+\log(x+1)-\log7=\log2x-\log x \)
\( ?=x \)
\( \log_89-\log_83+\log_4x^2=\log_81.5+\log_82+\log_4(-x^2-11x-9) \)
?=x
\( \frac{2\ln4}{\ln5}+\frac{1}{\log_{(x^2+8)}5}=\log_5(7x^2+9x) \)
\( x=\text{?} \)
\( \log4x+\log2-\log9=\log_24 \)
?=x
\( \log_9e^3\times(\log_224-\log_28)(\ln8+\ln2) \)
Defined domain
x>0
x+1>0
x>-1
We reduce by: and by
Undefined domain x>0
Defined domain
?=x
To solve the equation: , we proceed as follows:
Step 1: Simplify Both Sides
On the left-hand side (LHS), apply logarithmic subtraction:
.
Note remains and convert using the base switch to :
.
Thus, the LHS combines into:
(because ).
On the right-hand side (RHS):
Combine:
.
Also apply for term:
.
Step 2: Equalize Both Sides
Equate LHS and RHS logarithmic expressions:
.
The cancels out on both sides, leaving:
.
Step 3: Solve for
Since the denominators are equal, set the numerators equal:
.
Translate this into an exponential equation:
or
.
Let , solve the resulting quadratic equation:
.
Then, finding valid by allowing roots of polynomial calculations should yield laws consistency:
or rather substituting potential values. After appropriate checks:
The valid that satisfies the problem is thus .
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
?=x
\( \frac{\log_45+\log_42}{3\log_42}= \)
\( \log_64\times\log_9x=(\log_6x^2-\log_6x)(\log_92.5+\log_91.6) \)
Calculate the value of the following expression:
\( \ln4\times(\log_7x^7-\log_7x^4-\log_7x^3+\log_2y^4-\log_2y^3-\log_2y) \)
\( \frac{2\log_78}{\log_74}+\frac{1}{\log_43}\times\log_29= \)
\( \frac{\log_311}{\log_34}+\frac{1}{\ln3}\cdot2\log3= \)
For all 0 < x
Calculate the value of the following expression:
\( \frac{\log_76-\log_71.5}{3\log_72}\cdot\frac{1}{\log_{\sqrt{8}}2}= \)
\( -3(\frac{\ln4}{\ln5}-\log_57+\frac{1}{\log_65})= \)
\( \frac{1}{\ln4}\cdot\frac{1}{\log_810}= \)
\( \log_3x^2\log_527-\log_58=\ln e \)
\( \log_23x\times\log_58=\log_5a+\log_52a \)
Given a>0 , express X by a
Given a>0 , express X by a
Find X
\( \ln8x\times\log_7e^2=2(\log_78+\log_7x^2-\log_7x) \)
Solve for X:
\( \ln x+\ln(x+1)-\ln2=3 \)
\( \frac{\log_8x^3}{\log_8x^{1.5}}+\frac{1}{\log_{49}x}\times\log_7x^5= \)
\( \log_x16\times\frac{\ln7-\ln x}{\ln4}-\log_x49= \)
\( \frac{\log_47\times\log_{\frac{1}{49}}a}{c\log_4b}= \)
Find X
For all x>0
Solve for X: