Unit 2 Logarithm help please , any help appreciated

[Zahta]

Can you guys please help me with these question i really appreciate it , if you did step by step.


solve 25^2x-5=625^8-3x


Solve49^5x/7^x-4


simplify 5log₁₀x-2log₁₀x³+1/2log₁₀x¹⁶

solve log₁₀2+5log₁₀x-log₁₀5-3log₁₀x=log₁₀40 for x

solve log₃(1/6x-2)^-1=4 for x

[Bismuth]

You will learn better if you take a look at your log laws and read off some examples your textbook provides rather than ask people for step by step solutions. It'll benefit fit you in the long run.

For example, for the first two, you would want the bases to be the same before you could apply any index laws.

[matrix]

for the first one, write the expression with the base of 5
(5²)^(2x-5)=(5⁴)^(8-3x)
multiply powers
equate exponents
solve for x

the second one is similar, but the base is 7.

[Zahta]

okay the last 3 i tried like 3 times and im getting wrong can you guys please help i can do it except the last steps

[matrix]

for log expressions
bring the numbers in front of log(x) as a power of x, eg 5log(x)=log(x⁵)
so log(x⁵)-log((x³)²)+log(x¹⁶)^1/2
simplify by multiplying indices
next apply the rule log(AxB)=log(A)+log(B) and log(A/B)=log(A) - log(B)
so, log(x⁵.x⁸/x⁶)=log (x⁷)=7log(x)

[Zahta]

thankyou so much i really appreciate it , 

Can i ask you two more questions and their abit different to those another ones


Sove for each of the following for x,
okay the teacher explained it but still did dont get it she said something about making it y for this question


the question is 3^x93^x-270=0
& 2^2x-2^x+1

[Lasercookie]

I'm going to assume they were from two separate questions. I'll look at just the second one.


The idea with these ones is that you convert them to a quadratic, and thus can factorise them easily and that allows you to solve for x easily.

e.g.


(note that and )

Factorise:


Null factor law:

and

so: and
For the first case, it would be undefined (ignore it? - would this undefined answer be considered an 'extraneous solution'), and for the second one, x = 1.

So the answer is x=1.

You can also confirm this by graphing the original function.

edit: khan academy on extraneous solution if you haven't come across the idea: http://www.youtube.com/watch?v=711pdW8TbbY (I don't know why this isn't mentioned in the textbooks :|)

[matrix]

3^x.93^x-270=0
(3 . 93)^x=270
apply log to both sides
log(3 . 93)^x=log(270)
x . log (279)=log(270)

x=log(270)/log (279) this is the ecaxt answer,
use calculator if approx value is required