Solve (10^2x)^2 = 3^(x+1)

Question

xapproachesinfinity · Answer

hmm is that
$$(10^{2x})^2=3^{x+1}$$

anonymous · Answer

yass

anonymous · Answer

@xapproachesinfinity

xapproachesinfinity · Answer

ok
so $$100^{x^2}=3^{x+1}$$
we use log base 3 or loge base 100
$$x^2=\log_{100}(3^{x+1})$$

xapproachesinfinity · Answer

we then use change of base
$$x^2=\frac{\log_3(3^{x+1})}{\log_3(100)}$$

xapproachesinfinity · Answer

can you continue>>>

anonymous · Answer

i dont get how you got log base 100?

xapproachesinfinity · Answer

i just applied log base 100 to get rid  of 100 to power x^2 in the left

xapproachesinfinity · Answer

after i used log property. change of base

anonymous · Answer

ok so now what do you do?

campbell_st · Answer

ok... so start by using the index law for power of a power which means multiplying the powers so you get

$$10^{4x} = 3^{x + 1}$$

now take the base 10 log of both sides 

$$\log_{10}(10^{4x} )= \log_{10}(3^{x + 1})$$

next apply the log laws for powers... 

$$4x \log_{10}(10) = (x + 1)\log_{10}(3)$$

does that make sense so far..?

campbell_st · Answer

and in the next set you will need to know 

$$\log_{10}(10) = 1$$

anonymous · Answer

yass the first part makes sense,  but i got lost on what you just put

anonymous · Answer

oh wait nevermind i got it

anonymous · Answer

so now we have 4x= x+1(log3)

campbell_st · Answer

ok... when you take the log of the base... its equal to 1.... 

$$10 = 10^1~~so~~~\log_{10}(10^1) = 1$$

remember, using logs is about finding the power that when applied to the base gives the number needed...

as a further example 

$$\log_{10}(100) = \log_{10}(10^2) = 2$$

the base is 10, the power is 2 so 10^2 = 100 the starting number

campbell_st · Answer

so if you make the substitution 

$$\log_{10}(10) = 1$$

then you get

$$4x \times 1 = (x + 1)\log_{10}(3)$$

distribute the right hand side

$$4x = xlog_{10}(3) + \log_{10}(3)$$

is that ok...?

anonymous · Answer

yeah i got that

campbell_st · Answer

ok... now collect the like terms 

$$4x - x \log_{10}(3) = \log_{10}(3)$$

and factorise

$$x(4 - \log_{10}(3)) = \log_{10}(3)$$

does that make sense... and can you solve for x, from here...?

anonymous · Answer

yeah i get it thank yo so much, you would divide next right?

campbell_st · Answer

that's correct, by everything in the brackets on the left

anonymous · Answer

ok i got it thnx