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 _{5}^{6x+4} = Log _{2x49}

Should be simple...right?

Thanks

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.

Re: Log Change of Base Solution Please

Quote:

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 _{5}^{6x+4} = Log _{2x49}

Should be simple...right?

Thanks

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

Re: Log Change of Base Solution Please

doing the Left hand side I then get

Log _{2x} (6x+4)/Log _{2x} 5 = Log _{2x}49

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

Re: Log Change of Base Solution Please

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

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 can__not__ be solved for $\displaystyle x.$

From what course did this problem arise?

And who would assign such a frustrating problem?

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.