How Do You Solve Logarithmic Equations with Coefficients?

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To solve the logarithmic equation log6(2x-5) + 1 = log6(7x+10), exponentiate both sides by 6 to eliminate the logarithm. The equation simplifies to 6^(log6(2x-5) + 1) = 7x + 10, which can be rewritten using the property of exponents. This leads to the equation 6(2x-5) = 7x + 10. Solving for x results in x = 8, which has been verified as correct. The key challenge was managing the +1 term on the left side of the equation.
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Homework Statement



Solve:

log6(2x-5)+1=log6(7x+10)

Homework Equations



-None-

The Attempt at a Solution



I know we will need to exponentiate both side by 6 to get rid of the log, or either take another log.

I need help with getting rid of the +1 on the left side of the = sign because that is what is baffling me.
 
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Hint: 6^{1+a} = 6 \times 6^a, doesn't it?
 
log6(2x-5)+1=log6(7x+10)

6log6(2x-5)+1=6log6(7x+10)

6(2x-5)=7x+10

12x-30=7x+10

5x=40

x=8

I checked it and it worked, thanks for the help man!
 
darshanpatel said:

Homework Statement



Solve:

log6(2x-5)+1=log6(7x+10)

Homework Equations



-None-

The Attempt at a Solution



I know we will need to exponentiate both side by 6 to get rid of the log, or either take another log.


I need help with getting rid of the +1 on the left side of the = sign because that is what is baffling me.

log66 = ?

Chet
 

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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