1. ## Log Change of Base Solution Please

My daughter has the following question on her home work. Although I'm pretty good on all the calculus for the life of me I can not figure this out and am obviously doing some thing wrong. If you can help that would be great.

The question is:

Using the log change of base rule solve for x:

Log 56x+4 = Log 2x49

Should be simple...right?

Thanks

2. ## Re: Log Change of Base Solution Please

Hey tommly.

The change of base rule says that log_a(x) = log_b(x)/log_b(a) for bases a and b.

Hint: Try changing the LHS to log_2(x) given that it is log_5(x) by using this formula.

3. ## Re: Log Change of Base Solution Please

Originally Posted by tommiy
My daughter has the following question on her home work. Although I'm pretty good on all the calculus for the life of me I can not figure this out and am obviously doing some thing wrong. If you can help that would be great.

The question is:

Using the log change of base rule solve for x:

Log 56x+4 = Log 2x49

Should be simple...right?

Thanks
Is this the equation? It's a bit hard to read. $\displaystyle \displaystyle \log_5{(6x+4)} = \log_{2x}{(49)}$

4. ## Re: Log Change of Base Solution Please

doing the Left hand side I then get

Log 2x (6x+4)/Log 2x 5 = Log 2x49

That does not really help me.

I've tried changing the RHS to Log 5 and that does not help me either. I end up with a Log multiplying a Log both the the variable x in it......

obviously missing something

5. ## Re: Log Change of Base Solution Please

guess no one knows .... i'll continue to try

6. ## Re: Log Change of Base Solution Please

Hello, tommiy!

Your use of [SUB] and [SUP] is confusing.

$\displaystyle \begin{array}{ccc}\text{It could mean: }& \log_5(6x+4) \:=\:\log_{2x}49 & [1] \\ \\[-3mm] \text{Or maybe: }& \log(5^{6x+4}) \:=\:\log(2x^{49}) & [2] \end{array}$

If it is [1], we have: .$\displaystyle \log_5(6x+4) \:=\:\frac{\log_5(49)}{\log_5(2x)}$

. . which becomes: .$\displaystyle \log_5(2x)\cdot\log_5(6x+4) \:=\:\log_5(49)$

And I see no way to solve for $\displaystyle x.$

If it is [2], we have: .$\displaystyle (6x+4)\log5 \:=\:\log(2x^{49})$

This is a transcendental equation.
It cannot be solved for $\displaystyle x.$

From what course did this problem arise?
And who would assign such a frustrating problem?

7. ## Re: Log Change of Base Solution Please

Its number 1 and its from there standard text book for Year 12 in Australia. It actually has he answer of 3.5 which is correct but I can not see how you manage to achieve this.