Urgent help required

**AeroScizor** · February 25th 2010, 12:50 AM

Hi,
I have encountered some challenging sums when doing my revision for my math test. Please try to help me as I will be doing the test tomorrow.

Here are the sums:

1)Find the exact value of x if (3x)^lg3=(4x)^lg4.

2)Without using a calculator, evaluate (lg 5)^2 + lg 2 lg 50.

3) Solve x in terms of a given that log base a 5 + 2= log base a (x+a) + log base a (x-3a).

Help would be appreciated. Thanks in advance.

**Grandad** · February 25th 2010, 04:09 AM

Hello AeroScizor

Welcome to Math Help Forum!

Originally Posted by AeroScizor

Hi,
I have encountered some challenging sums when doing my revision for my math test. Please try to help me as I will be doing the test tomorrow.

Here are the sums:

1)Find the exact value of x if (3x)^lg3=(4x)^lg4.

2)Without using a calculator, evaluate (lg 5)^2 + lg 2 lg 50.

3) Solve x in terms of a given that log base a 5 + 2= log base a (x+a) + log base a (x-3a).

Help would be appreciated. Thanks in advance.

1) Take logs of both sides:

$\log3\log(3x)=\log4\log4x$

$\Rightarrow \log3(\log 3 + \log x)=\log4(\log4 + \log x)$

Solve this equation for $\log x$ , using the 'difference of two squares' to simplify the result.

Answer: $x = \frac{1}{12}$

2) Note that $\log 2 = \log 10 - \log 5 = 1 - \log 5$ , and that $\log50 = \log10+\log5 = 1+ \log 5$

So $\log2\log50 =$ ... (Use the difference of two squares again.)

Answer: $1$

I'll come back to (3) later, when I have time, if no-one else has answered.

Grandad

**Grandad** · February 25th 2010, 06:06 AM

Hello again AeroScizor

Originally Posted by AeroScizor

...
3) Solve x in terms of a given that log base a 5 + 2= log base a (x+a) + log base a (x-3a).

Help would be appreciated. Thanks in advance.

Write $2$ as $2\log_a(a)=\log_a(a^2)$ , and then:

$\log_a5+\log_a(a^2) = \log_a(x+a)+\log_a(x-3a)$

$\Rightarrow \log_a(5a^2)=\log_a(x+a)(x-3a)$

$\Rightarrow 5a^2 = ...$

Simplify, factorise and solve the quadratic in $x$ . Not both roots are valid though, because $a,\; (x+a)$ and $(x-3a)$ must all be positive.

Answer: $x = 4a$

Can you complete it now?

Grandad

**AeroScizor** · February 26th 2010, 05:22 PM

Originally Posted by Grandad

Hello again AeroScizorWrite $2$ as $2\log_a(a)=\log_a(a^2)$ , and then:

$\log_a5+\log_a(a^2) = \log_a(x+a)+\log_a(x-3a)$

$\Rightarrow \log_a(5a^2)=\log_a(x+a)(x-3a)$

$\Rightarrow 5a^2 = ...$

Simplify, factorise and solve the quadratic in $x$ . Not both roots are valid though, because $a,\; (x+a)$ and $(x-3a)$ must all be positive.

Answer: $x = 4a$

Can you complete it now?

Grandad

Yes, I understand now. Thanks a lot for your help! Cheers!

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